Let $D$ be a discrete valuation ring, possibly non-commutative, with uniformiser $\pi$, and let $Q=M_n(Q(D))$, where $Q(D)$ denotes the ring of quotients of $D$. Let $v$ be the extension of the $\pi$-adic filtration on $D$ to $Q$.
We will assume that $Q(D)$ has characteristic $p>0$.
Given a matrix $A\in Q$, the $p$-growth of $A$ need not be constant, i.e. it is possible that $v(A^{p^m})>p^m{v(A)}$ for some $m\in\mathbb{N}$.
We say that the $p$-growth of $A$ is eventually constant if for some $k\in\mathbb{N}$, $A^{p^k}$ has constant growth, i.e. $v(A^{p^{k+m}})=p^m{v(A^{p^k})}$ for all $m\in\mathbb{N}$.
My question is, after passing to $Q'=M_n(K)$ for some finite-dimensional extension $K$ of $Q(D)$, can we always conjugate $A$ by something invertible in $Q'$ to get a matrix with an eventually constant growth rate?
This is always possible if $D$ is commutative, because after passing to a finite extension of $Q(D)$, we can put $A$ into Jordan normal form, and raising this to high $p$-th powers will yield a diagonal matrix, which has constant growth.
But the Jordan normal form does not hold in a non-commutative setting, so this approach fails. Does anyone know of an alternative method that holds in a non-commutative setting, or can anyone think of any counterexample?