In the plane ($n=2$) the assertion is true, it is the Jordan-Schoenflies theorem. The full content of the theorem is false already for $n=3$, due to the Alexander horned sphere. However, the part that it contradicts is that the exterior component is not simply connected, but the interior component still is a topological ball.
The so called "Alexander trick" is tempting, but it assumes the sphere is the boundary of unit ball to begin with... Attempting to adapt the proof in a naive manner fails for non-convex cases.
So - my question is this: Let $M$ be an $n-1$ dimensional sub-manifold of $\mathbb{R}^n$ that is homeomorphic to $S^{n-1}$. By the Jordan-Brouwer separation theorem, $M$ separates $\mathbb{R}^n$ to two connected components. Is the interior (bounded) component a topological $n$-disc, i.e. homeomorphic to $D^n$, with $M$ as its boundary?
(If it helps - smoothness may be assumed)
For the tame case you want the generalized Schönflies theorem of Brown and Masur.