I think this is equivalent to the smooth Schoenflies conjecture; the executive summary is that this is true because smooth balls (in any dimension) are isotopic to standard ones.
Here are some details, starting with some preliminary remarks. Your $\bar{B}^3$ is diffeomorphic to a closed 3-ball, so I'll take that as given. Its boundary is a 2-sphere in $\partial \bar{B}^4$, and so is isotopic to the standard 2-sphere in $S^3$ by the 3D Schoenflies theorem (Alexander's theorem). Also, the standard result about balls in a manifold being standard applies to a codimension $0$ ball $\bar{B}^n$ embedded in an $n$-manifold in a non-proper (or neat) way. In other words the boundary of $\bar{B}^n$ is divided into $\bar{B}^{n-1}_\pm$ with $\bar{B}^n \cap \partial M = \bar{B}^{n-1}_-$ lying in the boundary of $M$, and $\bar{B}^{n-1}_+$ properly embedded. You really should worry a bit about corners here, but I'm going to ignore this (and in what follows).
Suppose first that we have $\bar{B}^4$ as described and that the Schoenflies conjecture holds. By the remarks above, do a preliminary isotopy so the boundary is standard. Now add a 4-ball $B_0^4$, divided into balls $U^0_\pm$, where $\bar{U}^0_+ \cap \bar{U}^0_-$ is a standard $\bar{B}_0^3$. Then $\bar{B}_0^3 \cup \bar{B}^3$ is an embedded sphere, and assuming the Schoenflies theorem, bounds balls on either side. Each of these balls is of the form $U^0_\pm \cup U_\pm$. But since we know that $U^0_\pm$ is a ball, its embedding is standard, so it follows that $U_\pm$ is a ball as well.
The converse is also straightforward. Given $S^3 \subset S^4$, dividing $S^4$ into regions $A_+ \cup A_-$, remove a small ball $B^4_0$ centered at a point of $S^3$. The complement of $B^4_0$ is a ball divided into regions $U_\pm$ as in your description, and by hypothesis each of $U_\pm$ is a ball. It follows that $A_\pm$ are balls as well.
