Let $k$ be an algebraically closed field of characteristic $p>0$, and let $G$ be a reductive group defined over $\mathbb{F}_p$. For any $d\in\mathbb{Z}^+$, let $C_d(G)$ be the set of conjugacy classes of elementary abelian $p$-subgroups of $G(\mathbb{F}_{q})$ of maximal rank, where $q=p^d$.

Is $S(G)=\displaystyle\sup_{d\in\mathbb{Z}^+}\{|C_d(G)|\}$ finite?

A bit of motivation for my question:

For $n=2m$, the largest rank of an elementary abelian $p$-subgroup of $G=\mathrm{SL}_n(\mathbb{F}_q)$ is $dm^2$, and all such subgroups are conjugate to the upper right $m\times m$ block. In this case, we see that $|C_d(G)|=1$ for all $d$. For $n=2m+1\ge 5$, the largest rank of an elementary abelian $p$-subgroup of $G=\mathrm{SL}_n(\mathbb{F}_q)$ is $dm(m+1)$, and all such subgroups are conjugate to either the upper right $(m+1)\times m$ block or the upper right $m\times(m+1)$ block, so that $|C_d(G)|=2$ for all $d$.

In both examples, not only is $S(G)$ finite, but $|C_d(G)|$ is independent of $d$. Perhaps this stronger statement is true. If not, I'd be interested to see a counter-example, that is, a reductive group $G$ and positive integers $d$ and $d'$ such that $|C_d(G)|\ne C_{d'}(G)|$.

I am fairly certain that $S(G)$ is finite for all simple groups of type $A$, $B,$ $C,$ and $D$. I'm less certain, but hopeful, that $S(G)$ is finite for the exceptional simple groups, and I have very little idea of what is happening in the general reductive case.