The fixed subalgebra of a finitely generated algebra

The condition on the characteristic is not necessary (nor is the condition that $k$ be a field, though it should be noetherian); see for instance Corollary 1.19 of http://people.fas.harvard.edu/~amathew/chgraded.pdf

Essentially, the point is that $A$ will be integral over $A^G$, and $A^G$ is "sandwiched" between $k, A^G, A$. From these facts, the result is not too complicated to prove.

The result that I only know under conditions on the characteristic is the strengthening to the case of $G$ a reductive algebraic group, acting (algebraically) on a finitely generated $k$-algebra: then $k$ is required to be of characteristic zero. The reason is that the proof uses semisimplicity of the category of $G$-representations, which is only true in characteristic zero.

Indeed the result is true with no restrictions on the characteristic: see $\S 14.6$, Theorem 339 of my commutative algebra notes. As I mention there, the result was first proved by Hilbert in characteristic zero and then in 1928 by Emmy Noether in arbitrary characteristic.