This occurs if and only if the matrices $Q^r$ and $Q^s$ are conjugate.
This is the case if and only if these matrices are conjugate over the
algebraic closure of $\mathbb{F}_p$. If $Q$ iis diagonalizable, then
things are straightfoward: if its eigenvalues are $\alpha_1,\ldots,\alpha_n$
then $Q^r$ is conjugate to $Q^s$ if and only if $\alpha_1^r,\ldots,\alpha_n^r$
are a permutation of $\alpha_1^s,\ldots,\alpha_n^s$.

An interesting case is where the characteristic polynomial of $Q$
is irreducible over $\mathbb{F}_q=\mathbb{F}_{p^k}$. In this case
the eigenvalues of $Q$ are $\alpha,\alpha^q,\alpha^{q^2},\ldots,\alpha^{q^{n-1}}$.
Then $Q^r$ and $Q^s$ are conjugate if and only if $s\equiv q^i r$
(mod $t$) for some $i$ with $0\le i < n$ and where $t$ is the multiplicative
order of $\alpha$ (and of $Q$).

When $Q$ is not diagonalizable, things get rather tedious. It's
not too bad if $Q$ is invertible and $r$ and $s$ are not divisible by $p$.
If $Q$ has a Jordan block of size $k$ with eigenvalue $\alpha$ then
$Q^r$ also has a Jordan block of size $k$ with eigenvalue $\alpha^r$
as long as $r$ is coprime to $p$. If $Q$ is singular or $r$ is a multiple
of $p$ then the Jordan block sizes of $Q^r$ might be different from those
of $Q$. Tedium ensues :-)