Here is a proof in the case where $p \not |m$ that doesn't involve modular representations, though it does involve a result of Gelfand and Kazhdan whose proof I can't remember, so I have no idea whether it's easy or hard.

A special case of their result is that two elements of $GL_n(\mathbb{Z}_p)$ are conjugate in that group if and only if they are conjugate in $GL_n(\mathbb{Q}_p)$.  Moreover, two elements of $GL_n(\mathbb{Q}_p)$ are conjugate in that group if and only if they are conjugate in $GL_n(E)$ for every extension $E/\mathbb{Q}_p$. Thus, if we assume that the reductions $\bar A$ and $\bar B$ of $A$ and $B$ are conjugate in $GL_n(\mathbb{F}_p)$, it will be enough to show that $A$ and $B$ are conjugate in $GL_n(E)$ for some $E$.  

Let $E/\mathbb{Q}_p$ be an extension containing all $m$th roots of unity.  These roots of unity, and thus the eigenvalues of $A$ and $B$, are completely determined by their images in the residue field of $E$, which are the eigenvalues of $\bar A$ and $\bar B$.  That is, $A$ and $B$ have the same multi-set of eigenvalues if and only if $\bar A$ and $\bar B$ do.

I believe that one could construct another proof from the result about $GL_n(\mathbb{Z}/p^n\mathbb{Z})$ mentioned in Gjergji's answer [at least if he really meant $GL_n(\mathbb{Z}_p/p^k\mathbb{Z})$], via some limiting process.