This is not true even if $A$ and $B$ are assumed to be commutative, for some trivial reasons. For example, let $A = k[t, s]/(s)(s-t)(s-2t)$, which represents three lines on a plane concurrent at the origin, and let $B = k[t, s]/(s)(s-t)\times k[t, s]/(s-2t)$, which is a partial resolution of $A$. There is an obvious (diagonal) morphism $A\to B$, which is injective and becomes bijective after inverting $t$. Set $C = k[t, s]/(s-t)(s-2t)$. We got a natural surjection $A\to C$ as well. Now put $M = C$ as an $A$-module. We have then $A_0 = A/tA\cong k[s]/(s^3)$, $B_0 = B/tB = k[s]/(s^2)\times k[s]/(s)$ and $M_0 = M/tM = k[s]/(s^2)$. Now the $A_0$-module $M_0$ extends uniquely into a $B_0$-module, which must be supported on the first factor of $B_0$, namely $k[s]/(s^2)$. However, the $A$-module $M$ does not extend to a $B$-module, since otherwise $M = (0, 1)M \oplus (1, 0)M$ would be decomposible, which would imply $M$ was decomposible as $A$-module, and then as $C$-module, contradicting the fact that $C$ is connected. I guess one should at least impose some connectivity assumption on the generic fibers. **response to the comment** Yes, I think. As I'm not quite familier with non-commutative algebras, there may be some errors in my arguments. For $x\in T=\mathrm{Spec}\;k[t]$, let us denote $N_x = N/(t-x)N$ for any $\mathcal{O}_T$-module $N$. For an open $U\subseteq T$, we denote $N_U = N\big|_U$. Regarded a coherent sheaf on $T$, the ring $\mathrm{End}_A(M)$ is a coherent subsheave of $\mathrm{End}_{k[t]}(M)$, which shows that $\mathrm{End}_A(M)$ is flat over $k[t]$. With the flatness, the new hypothesis $\dim_k \mathrm{End}_{A_0}(M_0) = 1$ implies that $\mathrm{rank}_{k[t]}\mathrm{End}_A(M) = 1$ and that $\mathrm{End}_{A_0}(M_0) \cong \left(\mathrm{End}_A(M)\right)_0$. For $x$ in some open neighbourhood $U\subseteq T$ of $0$, we have $\mathrm{End}_{A_x}(M_x) \cong \mathrm{End}_A(M)_x\cong k$ as well, which gives in turn $A_x / \mathrm{Ann}_{A_x}(M_x)\cong k$ and that $\mathrm{Ann}_{A_x}(M_x)\cong \mathrm{Ann}_{A}(M)_x$. As $A_x$ is finite dimensional over $k$, this implies that $M_x$ is a simple $A_x$-algebra for those $x\in U$. By the density theorem of Jacobson, we know that $A_x/\mathrm{Ann}_{A_x}(M_x) \cong \mathrm{End}_kM_x$. From now on, we will consider everything only over the good neighbourhood $U\subseteq T$. Let us just assume $U = T$ for notational convenience. Now set $I = \mathrm{Ann}_{A[t^{-1}]}M[t^{-1}]$, $I_{A} = I\cap A$ and $I_{B} = I\cap B$. We know that $I_A = \mathrm{Ann}_A(M)$ (the former containing the latter, the latter being maximal over each $x\in T$). The injection $i:A\to B$ induces $\bar i: A/I_A \to B/I_B$, which is again injective. Since $A/I_A$ and $B/I_B$ are the image of $A$ and $B$ in $A[t^{-1}]/I$, and since $A[t^{-1}]/I$ is flat over $k[t, t^{-1}]$, they are all flat over $k[t]$. The flatness implies that the induced $\bar i_0: (A/I_A)_0 \to (B/I_B)_0$ is injective. On the other hand, we know that $(A/I_A)_x$ and $(B/I_B)_x$ are of same dimension for all $x\in U$. We conclude that $\bar i_0$ is actually an isomorphism, and so is $\bar i$. Finally, the canonical epimorphism $B\to B/I_B$ composed with the isomorphism $B/I_B\cong A/I_A\cong \mathrm{End}_{k[t]}(M)$ will extend $M$ into a $B$-module. Well, it seems that the hypothesis of a unique extension of $M_0$ isn't quite necessary.