The case $n=4$ is open as far as I know.
The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk.
The case $n=5$ is equivalent to the smooth $4$-dimensional Poincare conjecture
(which is still open). Here is why:
Any homotopy $4$-sphere embeds smoothly into $\mathbb R^5$ (Sketch: homology $4$-sphere bounds a contractible smooth manifold $C$ [Kervaire, "Smooth homology spheres and their fundamental groups", Theorem 3]. In our case $\partial C$ is simply-connected, so attaching a collar on the boundary one gets a contractible $5$-manifold that is simply-connected at infinity, and hence it is diffeomorphic to $\mathbb R^5$ by a result of Stallings).
Any embedding of the standard $4$-sphere into $\mathbb R^5$ bounds a standard disk, see [Smale, "Differentiable and Combinatorial Structures on Manifolds", Corollary 1.3]. What Smale actually states is that any embedded $S^{n-1}$ in $\mathbb R^n$ bounds a standard disk unless $n=4$ or $7$. This was before he proved the h-cobodorsm theorem hence he excludes $7$.
Finally, as mentioned in comments if $n>5$, then $\bar D$ is diffeomorphic to the standard disk by the h-cobordism theorem (sketch: since by assumption $D$ is simply-connected at infinity, $\partial D$ is a homotopy sphere and $\bar D$ is a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one. Proving that this is an h-cobordism involves standard excision considerations in homology).
The situation is similar to the one of closed manifolds. One defines "boundary-prime" manifolds as those that cannot be decomposed nontrivially in a boundary-connected sum.
Note that if $M$ is connected, has nonempty boundary and is not prime, then $M$ is never boundary-prime. Namely, take a 2-sphere $S\subset M$ separating $M$ into two components none of which is a ball. Connect $S$ to $\partial M$ by a 1-handle $D^2\times [0,1]$ ($D^2\times \{0\}\subset S$, $D^2\times \{1\}\subset \partial M$). Now, remove from $S$ the open disk equal to $int(D^2)\times \{0\}$ and add to $S$ the annulus $\partial D^2\times [0,1]$. The resulting surface $S'$ is a 2-dimensional disk with $\partial S'\subset \partial M$. This disk will cut $M$ in two components, none of which is a ball.
This works no matter what $\partial M$ is, in particular, if $\partial M=T^2$.
Thus, in what follows (until the concluding paragraph), I will assume that $M$ is prime.
Lemma. If $\partial M$ is a torus, then $M$ is necessarily boundary-prime.
Proof. Let $D\subset M$ be a properly embedded disk splitting $M$ in two components, none of which is a 3-ball. (Splitting means that you remove from $M$ an open tubular neighborhood of $D$.) Since $D$ separates $M$, $\partial D$ separates $T^2=\partial M$, hence, bounds a disk $D'\subset T^2$. Taking the union $D\cup D'$ we obtain a non-properly embedded 2-sphere in $M$. Pushing the disk $D'$ slightly into $M$, we obtain a 2-sphere $S\subset M$ disjoint from the boundary. Since $D$ was splitting $M$ into two submanifolds none of which is a 3-ball, the same holds for $S$. Hence, $M$ is not prime, contradicting the standing assumption. qed
We continue the discussion of prime manifolds $M$ with toral boundary.
Such a manifold still can have compressible boundary. However, if this is the case, a boundary-compressing $D$ disk in $M$ is necessarily nonseparating. (Since every separating loop in the torus $T^2$ bounds a disk in $T^2$.)
Cutting $M$ open along $D$ results in a manifold $M'$ with spherical boundary. If $M'$ is homeomorphic to the 3-ball, then $M$ itself is a solid torus, $\hat{T}=D^2\times S^1$. Otherwise, attaching $B^3$ along $\partial M'$ results in a closed 3-manifold $N$ which is not $S^3$. Then $M= N\# \hat{T}$. But there is one more possibility, namely, $\partial M$ is incompressible. There are many manifolds like that, for instance, the exterior of any nontrivial knot in $S^3$.
Hence, we obtain the trichotomy for 3-manifolds $M$ which need not be $\partial$-prime.
To conclude: Suppose that $M$ is a connected (not necessarily prime) 3-manifold and $\partial M$ is homeomorphic to $T^2$. Then one of the following mutually exclusive properties holds:
$M$ is not prime, equivalently, is not $\partial$-prime.
$M=\hat{T}$.
$M$ is prime and $\partial M$ is incompressible.
Best Answer
I'll only discuss the first question (EDIT: Actually, I address the second question at the end). As Agol pointed out in the comments, for $n \geq 5$ this is an easy consequence of Newman's 1966 proof of the Poincare conjecture in the topological category.
I don't know if it was explicitly stated earlier than this. However, it can easily be derived from the main result of the paper
Brown, Morton, The monotone union of open (n)-cells is an open (n)-cell, Proc. Am. Math. Soc. 12, 812-814 (1961). ZBL0103.39305, MR0126835 (23 #A4129).
In fact, this works in all dimensions (including $3$ and $4$).
Brown's theorem is as follows. Assume that $M$ is a topological $n$-manifold and that for all compact $K \subset M$, there exists some open set $U \subset M$ with $K \subset U$ and $U \cong \mathbb{R}^n$. Then $M \cong \mathbb{R}^n$. Brown's proof is clever, but completely elementary.
To get the desired result from this, assume that $X = U_1 \cup U_2$ with $U_i \cong \mathbb{R}^n$ and that $X$ is compact. Let $\phi : \mathbb{R}^n \rightarrow U_1$ be a homeomorphism. It is enough to prove that $X \setminus \{\phi(0)\} \cong \mathbb{R}^n$. We will do this with Brown's theorem. Consider a compact set $K \subset X \setminus \{\phi(0)\}$. To verify Brown's criteria, it is enough to construct a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ such that $\psi(K) \subset U_2$.
For $r>0$, let $B(r) \subset \mathbb{R}^n$ be the ball of radius $r$. The set $U_1 \setminus U_2$ is compact, so there exists some $R>0$ such that $U_1 \setminus \phi(B(R)) \subset U_2$. Also, there exists some $\epsilon > 0$ such that $K \cap \phi(B(\epsilon)) = \emptyset$. It is easy to construct a homeomorphism $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $f(B(\epsilon)) = B(2R)$ and $f(0)=0$ and $f|_{\mathbb{R}^n \setminus B(3R)} = \text{id}$. We can therefore define a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ by $\psi(p) = \phi \circ f \circ \phi^{-1}(p)$ for $p \in U_1 \setminus \{\phi(0)\}$ and $\psi(p) = p$ for $p \notin U_1$. Clearly $\psi(K) \subset U_2$.
EDIT: Lee suggested that this might be able to address his second question too. I thought a bit about it, and I believe that it can. The key is the following "relative" version of Brown's theorem, which can be proven exactly like Brown's theorem.
Theorem : Let $(M,N)$ be a pair consisting of a topological $n$-manifold $M$ and a closed submanifold $N \subset M$. Assume that for all compact $K \subset M$, there exists some open set $U \subset M$ such that $K \subset U$ and such that the pair $(U,U \cap N)$ is homeomorphic to the pair $(\mathbb{R}^n,\mathbb{R}^{n-1})$ (the second embedded in the standard way). Then $(M,N) \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.
To apply this, assume that $X$ is a compact manifold with boundary and that $X = U_1 \cup U_2$ with $(U_i,\partial U_i) \cong (\mathbb{R}^n_{\geq 0},\mathbb{R}^{n-1})$. Double $X$ to get a closed manifold $Y$, and let $Y' \subset Y$ be the image of the boundary of $X$. The open sets $U_i$ double to give an open cover $Y = V_1 \cup V_2$. Letting $V_i' = V_i \cap Y'$, we have $(V_i,V_i') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$. Let $(M,M')$ be the result of deleting the image of $0$ in $(V_1,V_1')$. It is enough to prove that $(M,M') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$, and this can be proven just like above.
Of course, Agol answered the second question first -- it follows from the topological Schonfleiss theorem applied to the double, which was proven by Brown in
Brown, Morton, A proof of the generalized Schoenflies theorem, Bull. Am. Math. Soc. 66, 74-76 (1960). ZBL0132.20002, MR0117695 (22 #8470b).
Mazur had earlier proven a weaker result. This requires the sphere to be bicollared, but this holds. Indeed, from the assumptions the sphere is locally bicollared, and Brown proved in
Brown, Morton, Locally flat imbeddings of topological manifolds, Ann. Math. (2) 75, 331-341 (1962). ZBL0201.56202, MR0133812 (24 #A3637).
that this implies that the sphere is bicollared. See
Connelly, R., A new proof of Brown’s collaring theorem, Proc. Am. Math. Soc. 27, 180-182 (1971). ZBL0208.50704, MR0267588 (42 #2490).
for a super-easy proof of Brown's collaring theorem.