If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to 
$$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ 
where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. 
Thus, two homomorphisms $f,g\to M_n(k)$ are equivalent iff $ker{f}=ker{g}$.

**Claim:** Given a norm $\|.\|_A$ on $A$, there exists $\epsilon>0$ such that two homomorphisms $f,g:A\to M_n(k)$ are equivalent whenever $\|f(a)-g(a)\|\leq \epsilon\|a\|_A$ for all $a\in A$.

**Proof.** If $A$ is nilpotent, there is nothing to prove: every homomorphism $f\to M_n(k)$ is identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\epsilon = \frac{1}{2}\min\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\}.$$

Now suppose $f,g\to M_n(k)$ are non-equivalent homomorphisms. Then, either $p_i\subseteq ker{f}\backslash ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq ker{g}\backslash ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\epsilon\|p_i\|_A$.