[Update below:  I think this answers the question now.]  Let's suppose that we're working over an algebraically closed field $k$ of characteristic $p$.  Let $Fr$ denote the Frobenius automorphism.

From a recent paper of Tanaka and Kaneta (Hiroshima Math. J. 2008), I found that $Aut(G_a^n)$ (the automorphism group of the vector group $G_a^n$ over $k$) can be identified with:
$$\{ A \in M_n(k)[[F]]^\times : A,A^{-1} \in M_n(k)[F] \}.$$
Here, we work in a ring of noncommutative power series in the formal variable $F$, with coefficients in the matrix ring $M_n(k)$, subject to the natural relation:
$$F \cdot m = Fr(m) \cdot F.$$
I.e., one may pass all F's to the right, by applying Frobenius to the entries of the matrices.

Now, there is a natural surjective homomorphism $lin$ from $Aut(G_a^n)$ to $GL_n(k)$, obtained by taking the "constant term" of the power series.

What does the kernel of $lin$ look like?  Is it (representable by?) a pro-unipotent group scheme over $k$?  I have no idea.

Though it's a bit obvious, the "Refined Question" would be answered if one could prove the following:  If a group $G$ acts on the vector group $G_a^n$ via $\alpha: G \rightarrow Aut(G_a^n)$, then $Ker(lin \circ \alpha)$ is a unipotent subgroup of $G$.

**Update:**  Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

I think that this does it, though I'm shaky on pro-affine groups.  But the image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969.  Happily this result is right before the assumption of char=0).  Any reductive subgroup of a unipotent group is trivial, in this same reference.  Hence the image of a reductive group in $Ker(lin)$ must be trivial.

I think this answers the question, as long as I'm not missing any subtleties in the definitions of reductive/unipotent in the pro-affine setting.