Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasiprojective homogeneous Gvariety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$variety is already quasiprojective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical abstract isomorphism $\pi:X \rightarrow G/G _x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counterexamples or is any (quasiprojective) homogeneous $G$variety up to isomorphism of the form $G/H$?

It depends on what you mean by "closed subgroup". If you mean a Zariski closed subset which forms a subgroup then the answer is no. If you mean a closed subgroup scheme, then the answer is yes. An example where you need to use the second definition is the Frobenius map $F\colon G \to G^{(p)}$. If we let $G$ act on $G^{(p)}$ through $F$ then the action is transitive and indeed $G^{(p)}$ is isomorphic to $G/Ker F$. However, unless $G$ is zerodimensional $Ker F$ is a nontrivial finite group scheme whose $k$points consist of just the identity. Note however that $G/H$ is always quasiprojective even when $H$ is a subgroup scheme so all homogeneous $G$spaces are quasiprojective. 


An aside: BCnrd's argument above shows that, over any perfect field $k$, $X$ is isomorphic to $G/H$ if and only if $X$ has a $k$point. Otherwise, consider, e.g., conics with no point: they are homogeneous under a twist of $G=SO_3$ but are not $G/H$. I don't know what happens for more general base schemes. 

