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.

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.

EDIT: Here's a better partial answer, in the opposite direction. In set theory with atoms, BPI does not imply that every family of nonempty finite sets has a minimal blocking set. Specifically, in Mostowski's linearly ordered model, which is known to satisfy BPI (a theorem of Dan Halpern), the following family of finite, nonempty sets has no minimal blocker. Start with the disjoint union of countably many copies of the set $A$ of atoms, i.e., start with $\omega\times A$. Let $\mathcal G$ consist of all the $(n+1)$-element subsets of the $n$-th copy $\{n\}\times A$ for all $n\in\omega$. A minimal blocking set for this $\mathcal G$ would have to consist of (exactly) all but $n$ elements of the $n$-th copy, for all $n$. But no such set has finite support and therefore no such set is in the model.