$\DeclareMathOperator{\proj}{Proj}
\DeclareMathOperator{\grring}{\mathbf{GrRing}}
\DeclareMathOperator{\schpart}{\mathbf{Sch}_{part}}
\DeclareMathOperator{\sch}{\mathbf{Sch}}
\def\op{\mathrm{op}}
\DeclareMathOperator{\spec}{Spec}
\def\Q{Q}
\def\Fp{F}
\DeclareMathOperator{\id}{id}
\def\O{\mathcal{O}}$I will address the question in the title. First, contrary to popular belief:
$\proj$ is a functor.
Namely,
$$
\label{proj}\tag{1}
\proj:\grring^\op\to\schpart
$$
is a functor from the category of commutative unital graded rings to the category of schemes with partially defined morphisms of schemes. Similarly as the definition of the category of sets with partial functions (see here), we define $\schpart$ to be the category whose objects are schemes, and where a morphism $X\to Y$ in $\schpart$, that we call a partial morphism (of schemes) is a pair $(f,U)$ where $U\subset X$ is an open set and $f:U\to Y$ is a morphism of schemes. The composite
$$
X\xrightarrow{(f,U)}Y\xrightarrow{(g,V)}Z
$$
is defined as $(g\circ f_{|U\cap f^{-1}(V)},U\cap f^{-1}(V))$. The identity of $X$ is $(\operatorname{id}_X,X)$. One can check these specifications satisfy the axioms of a category.
Note that for any two schemes $X,Y$, there's always the empty morphism $e_{XY}=(e_Y,\varnothing):X\to Y$ between them, where $e_Y:\varnothing\to Y$.
The morphisms $e_{XY}$ are zero morphisms, and the empty scheme is a zero object (i.e., an initial and final object), see Wikipedia or nLab.
We have a faithful embedding $\sch\to\schpart$, which is a bijection on objects, and that sends $f:X\to Y$ to $(f,X):X\to Y$. In this sense, $\schpart$ may be viewed as a “enlargement” of $\sch$ obtained by adding extra morphisms between the already existing objects. Equivalently, $\sch$ is a wide subcategory of $\schpart$. Note that any isomorphism in $\schpart$ lies in $\sch$.
The functor $\proj:\grring^\op\to\schpart$ sends a graded ring $A$ to its projective spectrum $\proj A$. The underlying set of $\proj A$ are the homogeneous prime ideals of $A$ that do not contain the irrelevant ideal $A_+=\bigoplus_{n\geq 1}A_n$; we call call these the relevant homogeneous prime ideals. We next define the action of $\proj$ on a morphism $\varphi:A\to B$ of graded rings. Note that the inverse image of a (prime) homogeneous ideal of $B$ along $\varphi$ is (prime) homogeneous. However, given a prime homogeneous ideal $\mathfrak{q}\subset B$, $\mathfrak{q}$ being relevant does not imply $\varphi^{-1}(\mathfrak{q})\subset A$ is relevant. We can nonetheless define a morphism $r_\varphi:U(\varphi)\to\proj A$, where $U(\varphi)=\{\mathfrak{q}\in\proj B\mid\varphi^{-1}(\mathfrak{q})\text{ is relevant}\}$, which is open: the set $U(\varphi)$ is the complementary set in $\proj B$ to $V_+(\varphi(A_+))$. That is,
$$
U(\varphi)=\bigcup_{\substack{f\in A_+\\\text{homogeneous}}}D_+(\varphi(f)).
$$
On underlying topological spaces, $r_\varphi$ is defined to be the restriction of the map $\operatorname{Spec}\varphi:\operatorname{Spec}B\to\operatorname{Spec}A$ to $U(\varphi)$. The construction of $r_\varphi$ can be read in [SP, 01MY] or [EGA II, 2.8.1].
Therefore, we can define $\proj\varphi$ to be the partial morphism of schemes
$$
\proj\varphi=(r_\varphi,U(\varphi)).
$$
If $A\xrightarrow{\varphi}B\xrightarrow{\psi}C$ are morphisms of graded rings, then it is easy to see that $U(\psi\circ\varphi)\subset U(\psi)$. Thus [SP, 01MZ] is precisely the statement that \eqref{proj} is a functor, see also [EGA II, 2.8.4].
Once this fact has been stablished, we may ask, as OP did,
(OQ). Does $\proj$ have an adjoint?
Because of the contravariancy of $\proj$, this is more precisely phrased as
(Q). Does $\proj$ have a mutual right or mutual left adjoint? [R, Definition 4.3.1].
This question is motivated by the fact that the global sections functor is mutually right adjoint to the $\operatorname{Spec}$ functor,
$\operatorname{Hom}_{\mathbf{Ring}}(A,\Gamma(X,\mathcal{O}_X))
\cong\operatorname{Hom}_{\sch}(X,\spec A)$.
We can answer:
Proposition. $\proj$, as a functor \eqref{proj}, does not have a mutual right adjoint.
Proof. The category $\grring$ has pushouts and they are preserved by the forgetful functor $\grring\to\mathbf{Ring}$, see here or [M, Sect. 6] to see what the grading in $A\otimes_RB$ is. In particular, $\grring$ has coproducts (namely, $A\amalg B=A\otimes_{\mathbb{Z}}B$, for $\mathbb{Z}$ with the graduation $\mathbb{Z}_0=\mathbb{Z}$ is initial in $\grring$). Now, if $\proj$ had a mutual right adjoint, then it would need to send pushouts to pullbacks, and in particular coproducts to products. Fix a non-zero integer $n$, and let $Q$ be a ring where $n$ is a unit, and $F$ be a ring where $n=0$ (for instance, $Q=\mathbb{Q}$, $F=\mathbb{F}_p$, and $n=p$). We will show that
$$
\proj(\Q[x]\amalg_{\mathbf{Ring}}\Fp[x])\not\cong
\proj(\Q[x])\times_{\schpart}\proj(\Fp[x]).
$$
On the one hand,
$
\proj(\Q[x]\otimes_\mathbb{Z}\Fp[x])
=\varnothing,
$
for $\Q[x]\otimes_\mathbb{Z}\Fp[x]=0$ (if $f\in\Q[x]$ and $g\in\Fp[x]$, then $f\otimes g=n\frac{f}{n}\otimes g=\frac{f}{n}\otimes(ng)=0$).
On the other hand, $\proj(R[x])\cong \spec(R)$ (very generally proven in [EGA II, 3.1.7], but see also this), and in $\schpart$, the product of $\spec\Q$ with $\spec\Fp$ does exist. It is the scheme
$\spec\Q
\times_{\schpart}
\spec\Fp=\spec\Q\amalg\spec\Fp$, along with the projection morphisms
$$
\spec \Q
\xleftarrow{(\id,\spec\Q)}
\spec\Q\amalg\spec\Fp
\xrightarrow{(\id,\spec\Fp)}
\spec \Fp.
$$
Indeed, let $\spec\Q\xleftarrow{(f_Q,U_Q)}X\xrightarrow{(f_F,U_F)}\spec\Fp$ be partial morphisms. Then $U_Q\cap U_F=\varnothing$, for $n$ is a unit in $\O_X(U_Q)$, but it vanishes as a section in $\O_X(U_F)$. Thus we have a partial morphism $(f_Q,U_Q)\amalg(f_F,U_F)=(f,U):X\to\spec\Q\amalg\spec\Fp$. Conversely, any partial morphism $(f,U):X\to\spec\Q\amalg\spec\Fp$ is of this form. $\square$
This answers half of (Q). To answer the other half, i.e., showing that $\proj$ does not have a mutual left adjoint, one would follow a similar strategy: the category $\grring$ is complete and the forgetful functor $\grring\to\mathbf{Ring}$ is continuous (i.e., preserves all limits). One would then need to compute a limit in $\grring$ whose image under $\proj$ does not induce a colimit in $\schpart$. An easy case would be a product in $\grring$. Note that $\schpart$ has coproducts; actually the functor $\sch\to\schpart$ preserves coproducts.
I would study the scheme $\proj(\prod_{\mathbb{N}}R[x])$, and try to determine whether it is isomorphic to $\spec(\prod_{\mathbb{N}}R)$, for the latter is known to not be isomorphic to $\coprod_{\mathbb{N}}\spec(R)$.
References
[SP]. The Stacks Project Authors, The Stacks Project.
[EGA II]. Grothendieck, Alexandre; Dieudonné, Jean (1960). "Éléments de géométrie algébrique: II. Étude globale élémentaire de quelques classes de morphismes". Publications Mathématiques de l'IHÉS. English translation.
[R]. E. Riehl, Category Theory in Context.
[M]. D. Murfet, Graded Rings and Modules.
