I think it is true when we assume $M$ is a compact oriented Riemannian manifold.
Under this assumption, we can define an $L^2$-inner product on the space $\Omega^k(M;\mathrm{ad}P^r)$ by:
$$\begin{aligned}(\xi_1,\xi_2)_{L^2}&:=\int_M\langle\xi_1,\xi_2\rangle_x\mathrm{vol_M}\\&=\int_M\xi_1\wedge *\xi_2\end{aligned}$$
where $\mathrm{vol}_M$ is Riemannian volume form on $M$, $*$ is the Hodge *-operator and $\langle\cdot,\cdot\rangle_x$ is the fiber-wise $\mathrm{Ad}G^r-$invariant inner product on $\mathrm{ad}P^r$.
In particular, we can identify the Lie algebra $\mathrm{Lie}(\mathcal{G}^r)\cong\Omega^0(\mathrm{ad}P^r)$ with its dual, it will be convenient for us write a moment map.
Now, to proof my claim, I need to establish some computaional results at first.
Recall that a connection in $\mathscr{A}(P)$ is a $\mathfrak{g}-$valued 1-form defined on the principal bundle $\pi:P\longrightarrow M$, by identifying $$\mathscr{A}(P)=T^*\mathscr{A}(P^r)=\mathscr{A}(P^r)\times \Omega^1(M;\mathrm{ad}P^r)$$
we can write the connection $A$ as
$$A=a+(\pi^*\beta)i:=(a,\beta)$$
where $a$ is a connection on the subbundle $P^r$, $\beta\in\Omega^1(M;\mathrm{ad}P^r)$ is an $\mathrm{ad}P^r-$valued 1-form on $M$.
A tangent vector $\xi\in T_A\mathscr{A}\cong\Omega^1(M;\mathrm{ad}P)$ can be decoposed as
$$\xi=(\xi_1,\xi_2)\in T_a\mathscr{A}(P^r)\times\Omega^1(M;\mathrm{ad}P^r)=\Omega^1(M;\mathrm{ad}P^r)\times \Omega^1(M;\mathrm{ad}P^r)$$
where $\xi_1$ is simply $(dp_1)_A\xi$, and $p_1:A=(a,\beta)\mapsto a$ is the projection onto the 1st component.
Lemma 1: For $X\in\mathrm{Lie}(\mathcal{G}^r)$, the fundamental vector field $\underline{X}$ at a connection $A=(a,\beta)$ is $$\underline{X}(A)=(d_aX,0)\in T_a\mathscr{A}(P^r)\times\Omega^1(M;\mathrm{ad}P^r)$$where $d_a$ is the induced de Rham differential $\Omega^k(M;\mathrm{ad}P^r)\longrightarrow\Omega^{k+1}(M;\mathrm{ad}P^r)$.
The proof of this lemma is a very common argument in gauge theory. We shall notice that $\underline{X}(A)$ has only the first component is becasue the gauge group $\mathcal{G}^r$ acts by restriction of the total gauge group $\mathcal{G}(P)$, hence it acts by gauge transformation on $a$ but trivial on $\beta$.
Lemma 2: The Liouville 1-form $\alpha$ on $\mathscr{A}(P)$ is $$\alpha_A(\xi)=(\beta,\xi_1)_{L^2}$$ where $A=(a,\beta)$, $\xi_1=(dp_1)_A\xi$ is the first component of the tangent vector $\xi$.
This is can be computed directly by the definition of Liouville 1-form on cotangent bundle.
Lemma 3: For two vector fields $\xi,\eta$ on $\mathscr{A}(P)$, the value of $\xi(\alpha(\eta))$ at the connection $A=(a,\beta)$ is $$\alpha_A([\xi,\eta])+(\eta_1,\xi_2)_{L^2}$$
Proof: Let $\gamma(t)=(\gamma_1(t),\gamma_2(t))$ be the curve on $\mathscr{A}(P)$ such that
$$\xi\circ \gamma(t)=\gamma'(t),\quad \gamma(0)=A$$
then by definition and Leibniz law, we have
$$\begin{aligned}\xi(\alpha(\eta))(A)&=\left.\frac{d}{dt}\right|_{t=0}\left((dp_1)_{\gamma(t)}\eta(\gamma(t)),\gamma_2(t)\right)_{L^2}\\&=\alpha_A([\xi,\eta])+(\eta_1,\xi_2)_{L^2}\end{aligned}$$
. $\clubsuit$
Hence we can write the symplectic form immediately:
Corollary 4: The standard symplectic form on $\mathscr{A}(P)$ is $$\begin{aligned}\omega_A(\xi,\eta)&=(d\alpha)_A(\xi,\eta)\\&=\alpha_A([\xi,\eta])+(\eta_1,\xi_2)_{L^2}-(\eta_2,\xi_1)_{L^2}\end{aligned}$$ where $\xi,\eta\in T_A\mathscr{A}(P)$
With these results in mind, we can compute what is $\iota_{\underline{X}}\omega$. For $X\in\mathrm{Lie}(\mathcal{G}^r)$, and for an arbitrary vector field $\eta$ on $\mathscr{A}(P)$, we have at $A=(a,\beta)$ that
$$\begin{aligned}(\iota_{\underline{X}}\omega)_A(\eta)&=\omega_A((d_aX,0),\eta)\\&=\alpha_A([d_aX,\eta])-(\eta_2,d_aX)_{L^2}\end{aligned}$$
Finally, I claim that the map
$$\begin{aligned}\mu:\mathscr{A}(P)&\longrightarrow\mathrm{Lie}(\mathcal{G}^r)\\A=(a,\beta)&\mapsto d_a^*\beta\end{aligned}$$
is the moment map associated to the gauge action of $\mathcal{G}^r$. Where $d_a^*=-*\circ d_a\circ *$ is the Hodge co-differential.
It suffices to compute $((d\mu)_A\eta,X)_{L^2}$. Indeed, a parametrized curve $\gamma(t)$ in $\mathscr{A}(P)$ passing through $A$ with $\gamma'(0)=\eta$ is $$\begin{aligned}A+t\eta&=a+(\pi^*\beta)i+\pi^*(\eta_1+i\eta_2)t\\&=a+t\pi^*\eta_1+i\pi^*(\beta+\eta_2t)\end{aligned}$$ hence by the formal adjointness of the Hodge co-differential and Leibniz law, we have
$$\begin{aligned}((d\mu)_A\eta,X)_{L^2}&=\left(\left.\frac{d}{dt}\right|_{t=0}d^*_{a+t\pi^*\eta_1}(\beta+\eta_2t),X\right)_{L^2}\\&=\left.\frac{d}{dt}\right|_{t=0}\left(\beta+\eta_2t,d_{a+t\pi^*\eta_1}X\right)_{L^2}\\&=(\eta_2,d_aX)_{L^2}+\left(\beta,\left.\frac{d}{dt}\right|_{t=0}\underline{X}(\gamma(t))\right)_{L^2}\\&=(\eta_2,d_aX)_{L^2}-\alpha_A([d_aX,\eta])\\&=-(\iota_\underline{X}\omega)_A(\eta)\end{aligned}$$
It is indeed the moment map.
To sum up, we obtained the following result:
Theorem 5: For a complex Lie group $G$, and a principal $G-$bundle $P$ over a compact oriented Riemannian manifold $(M,g)$, the restriction action of a real gauge group $\mathcal{G}^r$ on the affine space of connections $\mathscr{A}(P)$ is Hamiltonian. The moment map is given by
$$\mu(A)=\mu(a+(\pi^*\beta)i)=d_a^*\beta$$
