If $K$ is an algebraically closed field of characteristic $p>0$, then $K((t))$, the field of Laurent series with coefficients in $K$, has infinitely many Galois extensions of degree $p$. Indeed, since the field $F_{p^n}$ with $p^n$ elements is contained in $K$, the Galois group of the polynomial $X^{p^n}-X=t^{-1}$ is isomorphic to the additive group of $F_{p^n}$, i.e. to $(\mathbb{Z}/p\mathbb{Z})^n$. Therefore the absolute Galois group of $K((t))$ is not finitely generated. Note that this argument only uses the fact that $K$ contains $\mathbb{F}_{p^n}$ for infinitely many $n$'s.

My question is whether the same holds true for arbitrary $K$ of positive charactersitic (perhaps we need to further assume that $K$ is infinite?).

Note that if $K$ is of characteristic $0$, then the restriction map $r\colon {\rm Gal}(K((t))) \to {\rm Gal}(K)$ of the absolute Galois groups is surjective and split. If the same would have hold for $K$ of positive characteristic, we would in particular have that $K((t))$ has infinitely many extensions of degree $p$ (not necessarily Galois) and hence ${\rm Gal}(K((t)))$ is not finitely generated.

However, I think that $r$ may not split in positive characteristic, since the proof that $r$ splits in characteristic zero is based on the fact that $\mathbb{A}^1_K$ is simply connected, and this fact fails if ${\rm char}(K)>0$.