I think this will work. There are a finite number of orbits of the action of $G$
on $X$ precisely when there is an open orbit and a finite number of orbits on
the complement of the orbit. Hence, it is enough to show that if there is an
open orbit of a point over the smaller field precisely when there is an open
orbit over the larger field. One direction is clear so it is enough to show that
there is an open orbit over the larger field, there is one over the smaller
field. However, consider the closed subscheme $S:=\{(g,x)\in G\times X | gx=x\}$
and the projection $S\to X$ on the second variable. A point of $X$ has an open
orbit precisely when the fibre has the smallest possible dimension (when $X$ is
irreducible which we may assume). However, there is an open subset (defined over
the smaller field) of $X$ with fibres of minimal dimension.

<b>Addendum</b>: As for the bijection between the orbits this is proved the same way, we have an open orbit which gives one orbit over each field and then we use Noetherian induction.