Edit3 ! Here is a simple proof I learnt from Michel Brion which works in characteristic $0$ and which gives a hypersurface (not a hyperplane) as desired. Consider the quotient $\mathbb P^n/G$. It is a projective variety of dimension $n$, so it has a finite morphism to the projective space $\mathbb P^n$. Now consider the finite morphism obtained by composition $f: X\to \mathbb P^n\to \mathbb P^n/G\to \mathbb P^n.$ By Bertini in characteristic $0$, $f^{-1}(H)$ is smooth for some hyperplane in the last $\mathbb P^n$. This is a smooth $G$-invariant hypersurface section in $X\subseteq \mathbb P^n$.
Unfortunately this method doesn't work in positive characteristic.
No this is not always possible. Consider a $G$ acting on $\mathbb P$ with finitely many fixed hyperplanes (e.g. cyclic permutation of the coordinates). For $d>>1$, there exists a smooth hypersurface $X$ of degree $d$, tangent to each of these fixed hyperplanes (in the space of hypersurfaces of degree $d$, to be tangent to a given hyperplane imposes two conditions, so it is enough to take $d$ bigger than twice the number of fixed hyperplanes).
Let $H$ be such that $H\cap X$ is smooth and stable by $G$. Then for any $g\in G$, $g(H)$ and $H$ both contain $H\cap X$ which is a smooth hypersurface of degree $d$ in $H$ and in $g(H)$. In particular $H\cap X$ is not linear. This implies that $g(H)=H$ because otherwise $g(H)\cap H=H\cap X$ by comparing the dimensions. But by constructio $H$ is then tangent to $X$, so $H\cap X$ is not smooth. Contradiction.
Edit It is more reasonable to ask the existence of $H$ such that $g(H)\cap X$ is smooth for all $g\in G$.
Edit 2 I have to withdraw my "proof" because my reasoning on the existence of smooth hypersurfaces $X$ tangent to given hyperplanes is not complete, moreover $X$ must also be stable by $G$. But my feeling is still that the answer should be negative in general because it can happen that there are only finitely many invariant hyperplanes.