Does $\partial A$ determine $A$?

Being bounded is not really an obstacle to your sphere example. You just need to tweak it a bit:

$$A_1=\big\{v\in\mathbb{R}^n\ \big|\ 1<\lVert v\rVert< 2\big\}$$ $$A_2=A_1\cup \big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert< 1\big\}$$ $$\partial A_1=\partial A_2=\big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert=1 \big\}\cup\big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert=2 \big\}$$

I think the other examples given so far are not correct. I think this works:

Take two disks in $\Bbb{R}^2$, and let one set be the union of one disk and the circle around the other, and let the other set be the union of the other disk with the circle around the former. Then they are closed, bounded, different than their boundaries, but have the same boundary while being different than one another.

I'll write a formula if this isn't clear.

EDIT. In fact, here's a much simpler example, in $\Bbb{R}$. Let $A_1=\{0,1\}\cup [2,3]$ and $A_2=[0,1]\cup \{2,3\}$.

EDIT 2. Like feynhat said, if you want a connected example, take my first example (in $\Bbb{R}^2$) and set $A_1=\{(x,y):x^2+y^2\leq 1\}\cup\{(x,y):((x-2)^2+y^2= 1\}$, $A_2=\{(x,y):x^2+y^2= 1\}\cup\{(x,y):((x-2)^2+y^2\leq1\}$, or something like this. The idea is to take two tangent disks. Perhaps someone can draw a figure and add it.

Consider a torus $\mathbb{T}^2$ in $\mathbb{R}^3$ and a circle $S^1$ on it such that it is null homotopic in $\mathbb{T}^2$. Then you get a disc on one side of the circle and a surface with nontrivial fundamental group on the other which share the same border. So the answer to the question is negative.