Edit: This answer is valid only in characteristic 0!
Yes. For a unipotent group $H$ the orbits of $H(k)$ in $X(k)$ coinside with the intersections of the form $H(\bar{k}).x \cap X(k)$, hence even Zarizki closed. In general, the orbits of $H(k)$ on sets of the form $H(\bar{k}).x \cap X(k)$ are classified by the first cohomology $H^1(k,H_x)$ for $H_x$ the stabilizer of $x$ in $H$. But Galois cohomology of unipotent groups vanish. This is a generalization of the fact that $H^1(k,\mathbb{G}_a) = \{1\}$ and in fact follows from this, since every unipotent group has a filtration with associated graded consist of several copies of $\mathbb{G}_a$, so by induction and some long exact sequences it follows.