No.

Here's a cheap argument. Let $G = PGL_n $ for $n>8$ even. Inside $G_{\mathbb F_2}$, we have the group of upper-triangular matrices which differ from the identity matrix only in the upper-right quadrant of the matrix. This group is isomorphic to $(\mathbb Z/2)^{n^2/4}$.

It suffices to prove that $(\mathbb Z/2)^{n^2/4}$ does not embed into $PGL_n$ in characteristic $0$.

Let's first lift from $PGL_n$ to $GL_n$. The commutator map from $(\mathbb Z/2)^{n^2/4} \times (\mathbb Z/2)^{n^2/4}$ to the $n$th roots of unity clearly lies in the $2$nd root of unity, and is symplectic, so it must have an isotropic subspace of dimension $n^2/8$. The inverse image of that group is abelian, thus also contains a subgroup isomorphic to $(\mathbb Z/2)^{n^2/8}$.

But any finite abelian subgroup of $GL_n$ may be simultaneously diagonalized, so $(\mathbb Z/2)^m$ only embeds in $GL_n$ if $m \leq n$. Since $n>8$ we have $n^2/8 > n$ and this is impossible.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\PGL{PGL}$No.

Here's a cheap argument. Let $G = \PGL_n $ for $n>8$ even. Inside $G_{\mathbb F_2}$, we have the group of upper-triangular matrices which differ from the identity matrix only in the upper-right quadrant of the matrix. This group is isomorphic to $(\mathbb Z/2)^{n^2/4}$.

It suffices to prove that $(\mathbb Z/2)^{n^2/4}$ does not embed into $\PGL_n$ in characteristic $0$.

Let's first lift from $\PGL_n$ to $\GL_n$. The commutator map from $(\mathbb Z/2)^{n^2/4} \times (\mathbb Z/2)^{n^2/4}$ to the $n$th roots of unity clearly lies in the $2$nd root of unity, and is symplectic, so it must have an isotropic subspace of dimension $n^2/8$. The inverse image of that group is abelian, thus also contains a subgroup isomorphic to $(\mathbb Z/2)^{n^2/8}$.

But any finite abelian subgroup of $\GL_n$ may be simultaneously diagonalized, so $(\mathbb Z/2)^m$ only embeds in $\GL_n$ if $m \leq n$. Since $n>8$ we have $n^2/8 > n$ and this is impossible.