Bounding the union of conjugates of a maximal subgroup of the Symplectic group over a finite field

Let $g \geq 1$ be a positive integer, and let $p$ be a prime. Consider the symplectic group $G := \operatorname{Sp}_{2g}(\mathbb{F}_p)$ of symplectic matrices with entries in $\mathbb{F}_p$. Let $M \subset G$ be a maximal subgroup and let $S = \bigcup_{h \in G} hMh^{-1}$.

Question: Is it true that $|S| = \alpha \cdot |G| + O(1/p)$, where $\alpha \in (0,1)$ is a constant, bounded away from $1$ and independent of $p$?

It's not too hard to see the answer is yes in the case where g = 1; see http://www.math.cornell.edu/~zywina/papers/Quantitative-HIT.pdf, Lemma 7.3. For higher g, I'm aware that there is a fairly nice classification of the maximal subgroups, such as that given in Kleidman and Liebeck, table 3.5.C. I think my question can be affirmatively answered for many of the geometric subgroup classes, by noting that the maximal subgroups are normalizers of certain finite index "classical subgroups" (as termed by Kleidman and Liebeck) and the complements of the classical subgroups in the maximal subgroups have trace 0, but I'm having difficulty showing this for all maximal subgroups, particularly for subgroups of Lie type.

Let $\Omega$ be the set of cosets of $M$, and consider the natural action of $G$ on $\Omega$. The set $S^C$ (the complement of $S$) is the set of derangements in this action.
So the upper bound you seek is equivalent to a lower bound on the proportion of derangements in the action of $G$ on $\Omega$. The lower bound you need is given by a much more general result of Fulman and Guralnick.
To state it, we need some notation: Let $H$ be a group acting on a set $X$. We write $\delta(H,X)$ for the proportion of elements in $H$ that are derangements in the action on $X$.
Theorem There exists a $\delta>0$ such that for any almost simple group $G$ and any set $\Omega$ on which $G$ acts transitively, $\delta(G,\Omega)>\delta$.