I learned the following fact from Bruhat and Tits's paper "Homomorphismes “abstraits” de groupes algebriques simples" Section 3.18 that

Let $k$ be a local field. Suppose that a $k$-group $H$ acts $k$-rationaly on a $k$-variety $M$ and $x$ is an element of $M(k)$ such that the map $h\mapsto hx$ for $h\in H$ of the group $H$ onto the orbit $Hx$ is separable . Then $H(k)x$ is locally closed in $M(k)$ with respect to metric Hausdorff topology.

Separable means its differential is surjective. I think any morphism is separable in characteristic 0. I have seen some examples for which the action is not separable, but all orbit are locally closed with respect to metric Hausdorff topology.

Q1: Is there a weaker (than separable) condition to ensure that all orbits are locally closed?

Q2: Do we know any examples for which there exists a non-locally closed orbit with respect to metric Hausdorff topology in positive characteristic case? It will be more interesting to have such action, which is also linear.

Thanks to user81562, I think we have the following:

If $F$ is a non-archimedean local field and $X$ algebraic variety defined over $F$. If $G$ is a linear algebraic $F$-group and $G\times X \rightarrow X$ is $F$-rational action. Then the action is constructive see Thm A on page 57 and in particular, all its orbits are locally closed (see proposition 6.8 c in http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf).