As stated, there is a non-semisimple counterexample: take $G$ to be the additive group $\mathbb{F}_q$ with a unipotent representation $\rho(a)=\left(\begin{matrix}1 & a\\ 0 & 1\end{matrix}\right)$. Its character is trivial but it is not defined over $\mathbb{F}_p$ because $GL_2(\mathbb{F}_p)$ does not have an order $p^2$ subgroup.

If we assume that $\rho$ is semi-simple then the answer is positive.

First, assume additionally that $\rho$ is absolutely irreducible. In general, let $L/K$ be a finite Galois extension and $\rho:G\to GL_n(L)$ be an irreducible representation with the characteristic polynomial of any element $\rho(g)$ defined over $K$. For any element $\sigma\in Gal(L/K)$ the representations $\sigma(\rho)$ and $\rho$ are semi-simple representations with equal characteristic polynomials, hence they are isomorphic(Bourbaki Algebra, Chapter 8, §20.6
Corollary 1 to thm 2) and the isomorphism is unique up to a scalar by the Schur's lemma. This gives an element $A(\sigma)\in PGL_n(L)$ for every $\sigma\in Gal(L/K)$ and by the unicity of the intertwining operators these matrices form a cocycle in $H^1(Gal(L/K),PGL_n(L))$. If this cocycle happens to be trivial, that is there is an element $B\in PGL_n(L)$ such that $A(\sigma)=\sigma(B)B^{-1}$, then $B^{-1}\rho B$ is a representation invariant under all the elements $\sigma$ and is the desired descent.

The group $H^1(Gal(L/K),PGL_n(L))$ injects into the Brauer group $Br(K)$. If $K$ is a finite field then $Br(K)$ vanishes and so does the group $H^1(Gal(L/K),PGL_n(L))$ which implies the vanishing of the obstruction to descending our representation.

If $\rho$ is not absolutely irreducible but is semi-simple(being semi-simple over $\mathbb{F}_q$ and over $\overline{\mathbb{F}_q}$ are equivalent conditions) then, after possibly enlarging $q$, it becomes isomorphic to a direct sum of two characters $\chi_1\oplus\chi_2$ such that $\chi_1(g)$ and $\chi_2(g)$ are roots of a quadratic polynomial with coefficients in $\mathbb{F}_p$. Thus theses characters must either be defined over $\mathbb{F}_p$ already or be defined over $\mathbb{F}_{p^2}$ such that $\overline{\chi_1}=\chi_2$ where $\overline{\bullet}$ is the non-trivial automorphism of this quadratic extension. The representation $\rho$ is then isomorphic to the base change of $G\xrightarrow{\chi_1}\mathbb{F}_{p^2}^{\times}\to GL_2(\mathbb{F}_p)$