The orientataion on $M$ would induces an orientation on $\partial M$. You can take a look at
Inducing orientations on boundary manifolds
An orientation on $\partial M$ is really a nowhere vanishing $n-1$-form $\omega$ on $\partial M$. This form satisfies $\int_{\partial M} \omega \neq 0$. By taking $-\omega$ if necessary, we can assume $\int_{\partial M} \omega >0$.
Now assume that $f : M \to \partial M$ exists. Then consider $i : \partial M \to M$ the inclusion. By condition on $f$, we have $f \circ i =id$ on $\partial M$. Let
$$\beta = f^*\omega$$
be a $n-1$ form on $M$. Then
$$\int_{\partial M} \omega = \int_{\partial M} (f\circ i)^*\omega = \int_{\partial M} i^*f^* \omega = \int_{\partial M} i^* \beta = \int_M d\beta$$
where the last step is Stokes theorem. But
$$d\beta= d f^* \omega = f^*d\omega = 0$$
(as $\omega$ is an $n-1$ form on $\partial M$, which has to be closed.)
Thus $\int_{\partial M} \omega= 0$ and it is a contradiction. Thus such an $f$ does not exist.
I think the point here is that an n-form which is non-zero everywhere defines a volume form and can be integrated to give a volume for the manifold which is positive. If you integrate an exact form you get zero, since there is no boundary. Hence the exact form cannot be a volume form.
Best Answer
Here's a proof which worked for both orientable and non-orientable manifolds. The proof comes from (my memory of) Milnor's book "Topology from the Differentiable Viewpoint". The proof only shows there is no smooth retraction. (With more work, one can show that there is a continuous retraction iff there is a smooth one, so this can be turned into a proof of the general statement.)
Suppose for a contradiction that $r:M\rightarrow \partial M$ is a retraction of a compact manifold $M$ onto its boundary. By Sard's theorem, there is a regular value $p\in \partial M$.
Then $r^{-1}(p)$ is a $1$-d submanifold in $M$. It is closed, being the inverse image of the closed set $\{p\}$, so is compact. A compact $1$-d manifold is homeomorphic to a disjoint union of a finite number of circles and closed, bounded intervals. In particular, the boundary of $r^{-1}(p)$ has even cardinality.
Now, the key point is that the boundary of $r^{-1}(p)$ must lie in $\partial M$. But $r^{-1}(p)\cap \partial M = \{p\}$ (since $r$ is a retraction), so is not of even cardinality.