The answer is no.

Consider any event $\omega$ such that $\xi_i(\omega) = 1$ for infinitely many $i$. Then for a given finite set $G$ one can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $1$ (to obtain $F$, simply add a lot of indices $i$ to $G$ for which $\xi_i(\omega) = 1$.

Similarly, if $\xi_i(\omega) = -1$ for infinitely many $i$, then for any given finite set $G$ we can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $-1$.

Hence, the net does not converge unless only the sequence $(\xi_i(\omega))$ is eventually constant. But the set of $\omega$ for which this happens has probability $0$.

The answer is no.

Consider any event $\omega$ such that $\xi_i(\omega) = 1$ for infinitely many $i$. Then for a given finite set $G$ one can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $1$ (to obtain $F$, simply add a lot of indices $i$ to $G$ for which $\xi_i(\omega) = 1$.

Similarly, if $\xi_i(\omega) = -1$ for infinitely many $i$, then for any given finite set $G$ we can find a finite superset $F$ such that $\frac{1}{\lvert F \rvert} \sum_{i \in F} \xi_i(\omega)$ is close to $-1$.

Hence, the net does not converge unless only the sequence $(\xi_i(\omega))$ is eventually constant. But the set of $\omega$ for which this happens has probability $0$.

I'm tempted to say that, intuitively speaking, a convergence property that relies in some way on harmonic analysis or maximal inequalities will, typically, depend on the order of the involved elements. (But experts in harmonic analysis might want to correct me if I'm oversimplifying things with this statement.)