$\DeclareMathOperator\Im{Im}$I don't think that smoothness of $X$ is necessary, just the fact that it is geometrically integral (I assume your varieties are geometrically integral). (As noted by Jason Starr, smoothness of $G$ is necessary.)

Here is a possible argument, please let me know if there are any mistakes.

As noted by AlexIvanov, it suffices to prove openness after passing to the algebraic closure of $k$, so we can assume that $k$ is algebraically closed.

Let $Z$ be the scheme theoretic closure of $G$ in $X$. The set theoretic image $\Im(G)$ of $G$ is open inside $Z$ by Chevalley's theorem and translating using the group action. After replacing $X$ with the $G$-stable open subvariety $X \setminus (Z \setminus \Im(G))$, we can assume that the natural map $G \to Z$ is surjective, in other words the orbit is closed. What we need to show at this point is that $Z = X$. We will argue by contradiction. Assume $Z\subset X$ is a proper closed subset.

We know that the tangent space of $Z$ at the base point $x$ agrees with that of $X$ (at $x$) by assumption. Using the fact that $k$ is algebraically closed, we can translate to see that the tangent space of $Z$ at any of its closed points coincides with that of $X$. But note that since $G$ is reduced (because $G$ is smooth), $Z$ is reduced. Since $k$ is algebraically closed, $Z$ is smooth over an open subset. In other words, the dimension of the tangent space of $Z$ for all points in an open dense agrees with the Krull dimension of $Z$. If $Z \subset X$ is a proper closed subset, the dimension of $Z$ is strictly smaller than that of $X$. This is a contradiction, because the dimension of the tangent space of $X$ at any closed point is at least the Krull dimension of $X$.