The issue is that $X(\mathbb{F}_p)/G$ is not the same thing as $(X/G)(\mathbb{F}_p)$. A simpler example is to take $p$ odd, $X = \mathbb{A}^1$ and let $S_2$ act by $\pm 1$. There are $\tfrac{p+1}{2}$ orbits, but the quotient space is $\mathbb{A}^1$ with the quotient map $x \mapsto x^2$. The quotient space has $p$ $\mathbb{F}_p$-points, but only $\tfrac{p+1}{2}$ of them (the squares) are in the image of $X(\mathbb{F}_p)$.

Points of $(X/G)(\mathbb{F}_p)$ index Frobenius stable $G$-orbits, so the other $\tfrac{p-1}{2}$ points correspond to the orbits of the form $\{ \pm x \}$ where $x^p = -x$ (other than the point $x=0$).

If you want to count $(X/G)(\mathbb{F}_p)$, there is a combined Burnside/Lefschetz formula. Choose $\ell$ relatively prime to $p$ and $|G|$, then a formula of Grothendieck tells us that $H^j(X/G, \mathbb{Q}_{\ell}) \cong H^j(X, \mathbb{Q}_{\ell})^G$, and this isomorphism is Frobenius equivariant. So the trace of Frobenius on $H^j(X/G, \mathbb{Q}_{\ell})$ is the same as the trace of Frobenius restricted to the subspace $H^j(X, \mathbb{Q}_{\ell})^G$. Now, the linear operator $\tfrac{1}{|G|} \sum_{g \in G} g$ on $H^j(X, \mathbb{Q}_{\ell})$ is an idempotent whose image is $H^j(X, \mathbb{Q}_{\ell})^G$. So the trace of Frobenius restricted to $H^j(X, \mathbb{Q}_{\ell})^G$ is the same as the trace of $\tfrac{1}{|G|} \sum_{g \in G} \text{Frob} \circ g$. So we get
$$\#((X/G)(\mathbb{F}_p)) = \tfrac{1}{|G|} \sum_{g \in G} \sum_j (-1)^j \text{Tr}{\big(}\text{Frob} \circ g : H^j(X, \mathbb{Q}_{\ell}) \longrightarrow H^j(X, \mathbb{Q}_{\ell}){\big)}.$$

I am not sure that there is a Burnside/Lefschetz style formula for $\#X(\mathbb{F}_p)/G$. Here is a troubling example: Take $p$ odd, $X = \mathbb{P}^1$ and let $S_2$ act by $[x:y] \mapsto [cy : x]$. Then $X(\mathbb{F}_p)$ has $p+1$ points. If $c$ is a quadratic residue, then there are two fixed points for $S_2$, namely $[\pm \sqrt{c}:1]$, otherwise there are none. So there are either $\tfrac{p+1}{2}$ or $\tfrac{p-1}{2}$ orbits for $S_2$ on $X(\mathbb{F}_p)$ depending on whether or not $c$ is a quadratic residue. I find it hard to imagine a Burnside/Lefschetz style formula which could take this information into account.

It might help to say that there is a purely topological version of this question. Let $G$ be a finite group and let $X$ be a compact topological space with an action of $G$ and also with an endomorphism $\phi$ that commutes with the $G$-action. (More generally, we could imagine that there is an automorphism $\sigma$ of $G$ with $\phi \circ g = \sigma(g) \circ \phi$.)

Then the two questions are how to count $(X/G)^{\phi}$ and how to count $(X^{\phi})/G$. These aren't the same thing: If there is a $G$-orbit which $\phi$ permutes nontrivially, then it will contribute to the first count but not the second.