Submersions are open maps; but the image of $M$ is compact in a Hausdorff space, and hence closed as well. So it's a clopen nonempty set. Since $\mathbf{R}^n$ is connected, it's the whole thing. But then $\mathbf{R}^n$ is the quotient of a compact space, so it's compact, which is not true.
In dimension 4, one has the following equivalence:
The set $\mathcal B_4$ of compact smooth manifolds homeomorphic to $B^4$, considered up to oriented diffeomorphism, is in canonical bijection with the set $\mathcal S_4$ of compact smooth manifolds homeomorphic to $S^4$, considered up to oriented diffeomorphism.
Proof. We construct maps both ways, which will be clearly inverse.
$C: \mathcal B_4 \to \mathcal S_4$. Pick $B \in \mathcal B_4$. By the resolution of the 3D Poincare conjecture, there is some oriented diffeomorphism $\varphi: \partial B \to S^3$. One may then "cap off the boundary": define $$C(B) = B \cup_\varphi B^4,$$ defined as $B \sqcup B^4$, modulo identifying $\partial B \cong S^3$ via the diffeomorphism $\varphi$.
$C$ is well-defined by Cerf's theorem that $\pi_0 \text{Diff}^+(S^3) = 1$: all diffeomorphisms are the same up to isotopy. Isotoping $\varphi$ above does not change the diffeomorphism type; so $C$ is a set map.
Conversely one has $D: \mathcal S_4 \to \mathcal B_4$, with $D(S)$ given by deleting the interior of some oriented embedding $\iota: B^4 \to S$. Oriented embeddings of balls into a connected manifold are unique up to isotopy (Palais, but straightforward); this is true in all dimensions.
Clearly $D(C(B)) = B$ (delete the ball you glued) and $C(D(S)) = S$ (glue in the ball you deleted). Therefore $C,D$ are inverse bijections.
However one has the following for $n \geq 6$.
The set $\mathcal B_n$ is trivial.
Proof: Pick $B \in \mathcal B_n$. Delete a standard ball from its interior; then we are provided with a compact manifold $W$, with $\partial W = S^{n-1} \sqcup \partial B$, so that $W \cup_{S^{n-1}} B^n = B$.
$W$ is an h-cobordism (algebra). Therefore by the h-cobordism theorem there is a diffeomorphism $W \cong S^{n-1} \times [0,1]$, which sends $S^{n-1}$ to $S^{n-1} \times \{0\}$ by the identity map. Therefore $W \cup_{S^{n-1}} B^n \cong B^n$. Therefore $B \cong B^n$.
Of course there are no exotic n-balls n = 1,2,3. For n=5 I do not know. I think the answer is that $\partial: \mathcal B_5 \to \mathcal S_4$ is an injection (maybe bijection?) but I don't know off the top of my head. Some googling or looking on Math Overflow should help. Obviously "exotic balls" will get you nowhere but some buzzwords like diffeomorphism might.
Best Answer
Yes, this is true. First, orient your $n$-manifold $M$ (your hypotheses imply that $M$ is contractible, so this is possible).
First, by your hypothesis, you obtain an increasing exhaustion $M_k \subset M$ of compact sets, diffeomorphic to the $n$-ball, so that each $M_k$ is contained in the interior of $M_{k+1}$.
This is enough; here is the idea. Write $B(k)$ for the unit ball of radius $k$ in $\Bbb R^n$. We may construct, for each $k$, some oriented diffeomorphism $\phi_k: M_k \to B(k)$. If we had $\phi_k \big|_{M_{k-1}} = \phi_{k-1}$, then by taking the increasing union of these $\phi_k$, we define a bijection $M \to \Bbb R^n$ which is a diffeomorphism on the interior of any compact set, and hence is a global diffeomorphism.
In practice, each successive $\phi_k$ has nothing to do with the previous one. Here is how we will resolve this.
Consider the map $g_k: \phi_{k+1}\phi^{-1}_k: B(k) \to B(k+1)$. All we know about this map is that it is a smooth oriented embedding into the interior.
This is a lemma of Cerf and Palais, independently; see here. The idea is to take any smooth embedding to the linear embedding in a chart given by the derivative at zero. In particular, we may find an ambient isotopy $f_t: B(k+1) \to B(k+1)$ which is the identity near the boundary, so that $f_0 = \text{Id}$ and $f_t g_k$ is a smooth isotopy between $f_0 g_k = g_k$ above and $f_1 g_k$ the canonical inclusion map.
Therefore, the map $f_1 \phi_{k+1}$ restricts to $\phi_k$ on $M_k$. So we choose this to be our given diffeomorphism $M_{k+1} \to B(k+1)$. Proceeding inductively, we have our desired result.