See Geometric Invariant Theory by Mumford, Fogarty, and Kirwan.
If $G$ is reductive, there will be a Zariski open subset $X^s$ of $X$ consisting of stable points for the action, for which the space of orbits $X^s/G$ exists as a variety, and the map $X^s \to X^s / G$ is regular. If the action is scheme-theoretically proper and free, this may be enough to guarantee that $X = X^s$. There is a lot of material in the above reference that is devoted to analysis of stability, so I am almost certain that this situation is treated in it.
When $G$ is not reductive, then extremely bad things can happen, as in Francesco's answer above.