Yes.

I'll show the expectation of $n$ minus the number of distinct eigenvalues is bounded.

This quantity is the sum over $\lambda$ in $\overline{\mathbb F_p}$ of the algebraic multiplicity of the eigenvalue $\lambda$ minus $1$.

A matrix $A$ gives $\mathbb F_p^n$ the structure of an $\mathbb F_p[A]$-module.

For each $\lambda \in \overline{\mathbb F_p}$ with minimal polynomial $f$, the algebraic multiplicity of $\lambda$ minus $1$ is at most the sum over $k \geq 2$ of the number of quotients of $\mathbb F_p^n$ isomorphic to $\mathbb F_p[A]/ f(A)^k$ plus the number of quotients of $\mathbb F_p^n$ isomorphic to $(\mathbb F_p[A]/ f(A) )^2$.

For each $\mathbb F_p[A]$-module $M$, the expectation of the number of quotients of $\mathbb F_p^n$ isomorphic to $M$ is equal to $$ \frac{ |\operatorname{Surj}_{\mathbb F_p} ( \mathbb F_p^n, M) |}{| \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) | | \operatorname{Aut}_{\mathbb F_p[A]} M |} \leq \frac{1}{ | \operatorname{Aut}_{\mathbb F_p[A]} M |}$$

since each quotient of $\mathbb F_p^n$ isomorphic to $M$ corresponds to $| \operatorname{Aut}_{\mathbb F_p[A]} M |$ surjections as $\mathbb F_p[A]$ modules from $\mathbb F_p^N$ to $M$ and each surjection as an $\mathbb F_p$-module is a surjection as an $\mathbb F_p[A]$-module with probability $\frac{1}{| \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) |}$.

Thus for $\lambda$ of degree $d$, the expectation of the multiplicity of $\lambda$ minus $1$ is at most

$$ \frac{1}{ |\operatorname{Aut} ( (\mathbb F_p[A]/ f(A) )^2) |} + \sum_{k=2}^{\infty} \frac{1}{ |\operatorname{Aut} (\mathbb F_p[A]/ f(A)^k ) |} = \frac{1}{ (q^{2d}-1) (q^{d}-d) }+ \sum_{k=2}^\infty \frac{1}{ q^{dk} - q^{d (k-1)}} $$ $$= O(\frac{1}{q^{2d}})$$

and summing this over all $\lambda$ of degree $d$ produces $O(q^{-d})$ which summing over $d$ is $O(1)$, as desired.

In fact $n$ minus the number of distinct eigenvalues even has a limiting distribution. In view of the above it suffices to check that for any finite set of eigenvalues the tuple of multiplicities of those eigenvalues has a joint limiting distribution, but this is known - it follows from the main result of this paper of Gilyoung Cheong and Myungjun Yu, though it's possible this was known earlier.

Yes.

I'll show the expectation of $n$ minus the number of distinct eigenvalues is bounded.

This quantity is the sum over $\lambda$ in $\overline{\mathbb F_p}$ of the algebraic multiplicity of the eigenvalue $\lambda$ minus $1$.

A matrix $A$ gives $\mathbb F_p^n$ the structure of an $\mathbb F_p[T]$-module, with $T$ acting by multiplication by $A$.

For each $\lambda \in \overline{\mathbb F_p}$ with minimal polynomial $f$, the algebraic multiplicity of $\lambda$ minus $1$ is at most the sum over $k \geq 2$ of the number of quotients of $\mathbb F_p^n$ isomorphic to $\mathbb F_p[T]/ f(T)^k$ plus the number of quotients of $\mathbb F_p^n$ isomorphic to $(\mathbb F_p[T]/ f(T) )^2$.

For each $\mathbb F_p[T]$-module $M$, a fundamental fact in the theory of random matrices is that a surjective $\mathbb F_p$-linear map $\mathbb F_p^n \to M$ is a $\mathbb F_p[T]$-homomorphism, with $T$ acting on $\mathbb F_p^n$ by multiplication by $A$, with probability $\frac{1}{| \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) |}$.

Indeed, an $\mathbb F_p$-module homomorphism $f \colon \mathbb F_p^n \to M$ if and only if the maps $x \mapsto f(A x)$ and $x\mapsto T f(x)$ from $\mathbb F_p^n$ to $M$ agree. If $f$ is surjective, we may choose bases for $\mathbb F_p^n$ and $M$ for which $f$ is projection to the first $m$ coordinates. In this basis $f(A e_i)$ may be calculated as the first $m$ coordinates of the $i$'th column of $A$ which take every possible value with equal probability and are independent of $f(A e_j)$ for $j\neq i$, so $x\mapsto f(A x)$ takes the value of each possible $\mathbb F_p$-homomorphism $\mathbb F_p^n\to M$ with equal probability, and in particular takes the value $x\mapsto T f(x)$ with probability $\frac{1}{ | \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) |}$.

From this, we can see that for an $\mathbb F_p[T]$-module $M$, the expectation of the number of quotients of $\mathbb F_p^n$ isomorphic to $M$ is equal to $$ \frac{ |\operatorname{Surj}_{\mathbb F_p} ( \mathbb F_p^n, M) |}{| \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) | | \operatorname{Aut}_{\mathbb F_p[T]} M |} \leq \frac{1}{ | \operatorname{Aut}_{\mathbb F_p[T]} M |}$$

since each quotient of $\mathbb F_p^n$ isomorphic to $M$ corresponds to $| \operatorname{Aut}_{\mathbb F_p[T]} M |$ surjections as $\mathbb F_p[T]$ modules from $\mathbb F_p^N$ to $M$ and each surjection as an $\mathbb F_p$-module is a surjection as an $\mathbb F_p[T]$-module with probability $\frac{1}{| \operatorname{Hom}_{\mathbb F_p}(\mathbb F_p^n, M) |}$.

Thus for $\lambda$ of degree $d$, the expectation of the multiplicity of $\lambda$ minus $1$ is at most

$$ \frac{1}{ |\operatorname{Aut} ( (\mathbb F_p[T]/ f(T) )^2) |} + \sum_{k=2}^{\infty} \frac{1}{ |\operatorname{Aut} (\mathbb F_p[T]/ f(T)^k ) |} = \frac{1}{ (q^{2d}-1) (q^{d}-d) }+ \sum_{k=2}^\infty \frac{1}{ q^{dk} - q^{d (k-1)}} $$ $$= O(\frac{1}{q^{2d}})$$

and summing this over all $\lambda$ of degree $d$ produces $O(q^{-d})$ which summing over $d$ is $O(1)$, as desired.

In fact $n$ minus the number of distinct eigenvalues even has a limiting distribution. In view of the above it suffices to check that for any finite set of eigenvalues the tuple of multiplicities of those eigenvalues has a joint limiting distribution, but this is known - it follows from the main result of this paper of Gilyoung Cheong and Myungjun Yu, though it's possible this was known earlier.