This is only a partial answer: If you assume (with notation as in the question) that every point is in only finitely many sets from $\mathcal G$, then the existence of a minimal blocking set follows from the Boolean prime ideal theorem (BPI).  One way to see this is via the compactness theorem for propositional logic (which is equivalent to BPI), applied to the following situation.  Let there be a propositional variable $\hat p$ for each point $p$ in the union of the family $\mathcal G$, and let $S$ be the set of the following sentences.  First, for each $F\in\mathcal G$, the disjunction of the $\hat p$'s for $p\in F$ is in $S$.  Second, $S$ contains, for each $p$, the implication
$$ \hat p\implies\bigvee_{F:p\in F\in\mathcal G}
\bigwedge_{q\in F-\{p\}}\neg\hat q$$
A truth assignment satisfying $S$ yields a minimal blocking set, namely the set of those $p$ whose $\hat p$ is assigned the value true.  (The displayed implication serves to ensure that, for each such $p$, there is a set in $\mathcal G$ containing it and no other element of the blocking set, so we get minimality.)  Every finite subset of $S$ is satisfiable, essentially because finite families have minimal blocking sets.  So compactness gives a satisfying assignment for $S$ and thus a minimal blocking set for the whole $\mathcal G$.

Unfortunately, if we drop the hypothesis that each $p$ is in only finitely many sets from $\mathcal G$, then this argument breaks down, because the disjunction in the displayed implication becomes an infinite disjunction, and there is no compactness theorem for such infinitary sentences.