Stable Conjugacy for Integer Matrices

Well, I think the answer is "no". Here's a construction: let R be the ring of integers of a real quadratic field K of class number > 1, let M be an invertible R-module of rank one which is not isomorphic to R, and let x be a fundamental unit in R. Then the action of x on M (viewed as a Z-module) determines a well-defined conjugacy class C_M in GL_2(Z), and similarly the action of x on R determines a conjugacy class C_R. I claim that these conjugacy classes are distinct, but become equal in GL_2(A).

They are distinct: indeed, M is recovered up to isomorphism from C_M since R identifies with the commutant algebra of x acting on the Z-module M.

They become equal in GL_2(A): in fact they become equal in GL_2 of the ring of integers in the Hilbert class field of K, since M and R become isomorphic there.

Here is an explicit realization of the counterexample suggested by Dustin. The field ${\mathbf Q}(\sqrt{10})$ has class number 2 and its Hilbert class field is obtained by adjoining $\sqrt{2}$. The ring of integers ${\mathbf Z}[\sqrt{10}]$ has (fundamental) unit $u:=3+\sqrt{10}$, whose minimal polynomial over ${\mathbf Q}$ is $T^2 - 6T - 1$. The two ideal classes in ${\mathbf Z}[\sqrt{10}]$ are represented by the ideals $(1)$ and $(2,\sqrt{10})$, which have ${\mathbf Z}$-bases $\{1,u\}$ and $\{2,\sqrt{10}\}$. Multiplication by $u$ on these two ideals is represented, using the indicated $\mathbf Z$-bases, by the respective matrices $A = (\begin{smallmatrix}0&1\\1&6\end{smallmatrix})$ and $B = (\begin{smallmatrix}3&5\\2&3\end{smallmatrix})$. These matrices are both in ${\rm GL}_2({\mathbf Z})$, they are not conjugate in this group, but they are conjugate by the matrix $U = (\begin{smallmatrix}\sqrt{2}&5+3\sqrt{2}\\1&3+2\sqrt{2}\end{smallmatrix})$, which lies in ${\rm GL}_2({\mathbf Z}[\sqrt{2}])$. That is, $UAU^{-1} = B$. This conjugating matrix $U$ has determinant $-1$. A matrix with determinant 1 and algebraic integer entries that satisfies $VAV^{-1} = B$ is $V = (\begin{smallmatrix}2\sqrt{2}&6\sqrt{2}+5\sqrt{3}\\\ \sqrt{3}&4\sqrt{2}+3\sqrt{3}\end{smallmatrix})$.

Quite generally, the matrix $M = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ satisfies $MA = BM$ if and only if $b=3a+5c$ and $d = 2a+3c$, and then $\det M = 2a^2 - 5c^2$. We can't solve $2a^2 - 5c^2 = \pm 1$ in ${\mathbf Z}$ (look at it mod 5), but we can solve it in ${\mathbf Z}[\sqrt{2}]$ using $a = \sqrt{2}$ and $c = 1$. That is how I found $U$. We can solve $2a^2 - 5c^2 = 1$ using $a = 2\sqrt{2}$ and $c = \sqrt{3}$, which is how I found $V$.