Note: a previous version of this answer had the extra assumption that $Y$ is compact. This version deals with the general case where $Y$ is possibly non-compact.
Yes, this is true. The proof is similar in spirit to the proof that
a map of non-zero degree between connected closed manifolds
of the same dimension is surjective.
$\newcommand{\bdry}{\partial}
\newcommand{\iso}{\cong}
\newcommand{\Z}{\Bbb{Z}}
\newcommand{Int}{\operatorname{Int}}$
Let $n$ be the dimension of the manifolds.
By assumption, $f$ induces a map of pairs $(X, \bdry X) \to (Y, \bdry Y)$.
As a first step, we will prove:
Proposition A: the map $H_n(X, \bdry X) \to H_n(Y, \bdry Y)$
induced by $f$ is non-zero.
Proof:
Consider the following portion of the long exact sequences of the pairs
with coefficients in $\Z_2$:
$$
\require{AMScd}
\begin{CD}
H_n(X) @>>> H_n(X, \bdry X) @>>> H_{n-1}( \bdry X)\\
{} @VVV @VVV \\
@. H_{n}(Y, \bdry Y) @>>> H_{n-1}(\bdry Y)
\end{CD}$$
where the vertical maps are induced by $f$.
We have the following observations:
The map $H_{n-1}(\bdry X) \to H_{n-1}(\bdry Y)$ is an isomorphism since $f$
restricts to a homeo from $\bdry X$ to $\bdry Y$.
$H_n(X) = 0$ since $X$ has the homotopy type of a non-compact manifold (see Proposition 3.29 of Hatcher). It follows that the boundary map $H_n(X, \bdry X) \to H_{n-1}(\bdry X)$ is injective.
$H_n(X, \bdry X) \iso \Z_2$, generated by the relative fundamental class $[X, \bdry X]$.
By inspecting the diagram, we deduce that the map
$f_*: H_n(X, \bdry X) \to H_n(Y, \bdry Y)$
is non-zero. $\square$
Now, suppose that $f$ is not surjective.
Then, there is a point in the interior of $Y$ which is not in the image of $f$.
Indeed, the image of $f$ is closed since $X$ is compact, so if the image contained $\Int(Y)$
it would also contain the closure of $\Int(Y)$, which is $Y$.
Pick then a point $y \in \Int(Y)$ which is not in the image of $f$.
Then, $f$ factors as follows:
$$
(X, \bdry X) \to (Y - \{ y \}, \bdry Y) \to (Y, \bdry Y)
$$
and we have a similar factorization of the map induced on homology.
We will now conclude using the following general fact
Proposition B: Let $Z$ be a connected $n$-manifold with boundary.
If $Z$ is non-compact, then $H_n(Z, \bdry Z) = 0$.
By applying this to $Z = Y - \{ y \}$, we obtain a contradiction, since we established in Proposition A that $f_*: H_n(X, \bdry X) \to H_n(Y, \bdry Y)$ is non-zero.
Note 1: If $Y$ is non-compact, we can already conclude by applying Proposition B to $Z = Y$,
so there is no need to consider $Y - \{ y \}$.
Note 2: as observed by Christophe Boilley in the comments, this proof still works
if we only assume that $f$ induces an embedding of $\bdry X$ inside $\bdry Y$.
Proof of Proposition B
I would be grateful if someone knows a reference for this, or a simpler proof.
We will use the following lemma, which can be found in the book Topology and Geometry by Bredon.
Lemma (Bredon, Corollary 7.11, p.346):
Let $M$ be an $n$-manifold without boundary and $A \subset M$ be a closed,
connected subset.
If $A$ is non-compact, then $H_n(M, M - A) = 0$.
Proof of Proposition B:
Let $U$ be a collar neighborhood of $\bdry Z$. Let $V = U - \bdry Z$.
By pushing $\bdry Z$ into the interior of $Z$ along the collar coordinate,
we get a map $(Z, \bdry Z) \to (\Int Z, V)$ which induces isomorphisms in relative homology.
Let $A = \Int Z - V$, which is a closed subset of $\Int Z$.
By choosing a smaller $U$ if necessary, $A$ will be homeomorphic to $Z$.
In particular, $A$ is connected and non-compact.
Therefore, we may apply the lemma (with $M = \Int Z$) and conclude:
$$
H_n(Z, \bdry Z) \iso H_n( \Int Z, V) = H_n(\Int Z, \Int Z - A) = 0.
$$
$\square$
