**Preamble:**

The original answer below is wrong. The mistake is that when restricting the open set $U$ to $U\cap X^{ps}$, the image of the GIT projection (call the map $\pi$) of $U$ may not equal the image of the quotient map $p:X^{ps}\to X^{ps}/G$. Certainly, $p(U\cap X^{ps})\subset \pi(U)$ but the point is that $U$ can contain some non-polystable points that get removed when intersecting and can change the topology of the image. This in fact was my original intuition with my attempted counter-example, but the example I choose was in fact open. So instead of trying another example, I decided to challenge my "gut" by trying to prove myself wrong and surprisingly came up with a "proof".

Anyway, *Friedrich Knop's answer is right.* I feel like I should delete this "answer," but on the other hand, sometimes failed attempts are instructive to others so I am not sure I will. I added a remark at the end that might be useful to the OP since the OP expressed interest in understanding the strong topology of $X//G$.

**Failed Counter Example:**

Any open orbit maps to a point, so generally the GIT quotient is not an open map (**see comments for the mistake**).

**Failed Proof of Openness:**

We work over $\mathbb{C}$. Take an open set $U\subset X$ then $U\cap X^{ps}$ is open in $X^{ps}$ (with respect to the relative topology) where $X^{ps}$ is the set of polystable points (points with closed orbits). Therefore $U\cap X^{ps}$ maps to an open set in $X^{ps}/G$ since $p:X^{ps}\to X^{ps}/G$ is an open map (this follows from the definition of the quotient topology and the fact that $G$ acts by homeomorphisms). But that set is equal to the image of $U$ under the GIT projection (**this step is the mistake**). Hence it is open in $X//G$ since $X//G\cong X^{ps}/G$ (see for example Theorem 2.1 here).

**Weak Correction:**

Whenever $p(U\cap X^{ps})\supset\pi(U)$ for all open $U,$ then $\pi$ is an open map in the strong topology. This follows from the above failed proof, since it fills the gap with an assumption.

**Remark:**

On the other hand, if one wants to understand the space $X//G$ in the strong topology one can replace $\pi:X\to X//G$ by $p:X^{ps}\to X^{ps}/G$. The latter is open while we now know the former might not be, but as a space in the strong topology $X//G$ remains homeomorphic to $X^{ps}/G.$ More still, as per Proposition 3.4 here, the usual quotient $X/G$ is homotopic to $X//G$.