Consider a smooth properly embedded surface $P\subset \mathbb{R}^3$. Then $\mathbb{R}^3= X\cup Y$, where $X\cap Y=P$ and $X, Y$ are properly embedded submanifolds with $\partial X=\partial Y=P$. By Mayer-Vietoris, we have an exact sequence $0=H_2(\mathbb{R}^3)\to H_1(P)\to H_1(X)\oplus H_1(Y)\to H_1(\mathbb{R}^3)=0$, so we see that $H_1(X)\oplus H_1(Y)\cong H_1(P) \neq 0$ unless $P$ is a union of smoothly properly embedded planes and spheres. Therefore at least one component of $\mathbb{R}^3\backslash P$ does not have perfect fundamental group, or else $P$ is a union of planes and spheres (since $P$ is smoothly properly embedded, there's a nice collar neighborhood, so $H_i(X)\cong H_i(int(X))$, and same for $Y$). In the case that $P$ is a union of properly embedded planes and 2-spheres, by Seifert-Van Kampen's theorem, each $\pi_1(X,x)$ and $\pi_1(Y,y)$, for $x\in X, y\in Y$ injects into $\pi_1(\mathbb{R}^3)$, so is trivial.

Let's apply this to your situation. I'll consider the case of $int(A)\backslash B$, since there's not issue of local connectivity for an open set. Then $P=\partial B \cap int(A)\subset int(A)\cong \mathbb{R}^3$ is a properly embedded smooth surface, so we have either there is a component of $int(A)\backslash B$ which does not have perfect fundamental group, or $P$ is a union of properly embedded planes in $int(A)$ (or a sphere), in which case each complementary region has trivial fundamental group. I think this answers at least one interpretation of your question.