I know that unconditional independence does not in general imply conditional independence, but does it make sense in this setting? We can assume $Z_i$ is discrete for each $i$.
Here, we can say $\{X_i| Z_i\}_i^n$ is independent iff $\{X_i | Z_i = z_i\}_i^n$ is independent for every possible realization of $Z_i$s, $\{z_i\}_i^n$, to avoid any "pseudo-random" variable/random random variable problems...
Let $A_1, A_2, ..., A_n$ be events in $\sigma(X_1), ..., \sigma(X_n)$, respectively, and let some $\{z_i\}_i^n$ be given. Suffices (but not necessary, I think...) to show that $$P\left(\bigcap A_i \middle| \bigcap \{Z_i = z_i\} \right) = \prod_i P\left(A_i \middle| Z_i = z_i\right) = \prod_i \frac{P\left(A_i \cap \{Z_i = z_i\}\right)}{P(Z_i = z_i)}.$$
Here is what I have so far: $$P\left(\bigcap A_i \middle| \bigcap \{Z_i = z_i\} \right) = \frac{P\left(\bigcap \{Z_i = z_i\} \middle| \bigcap A_i \right)P\left( \bigcap A_i\right)}{P\left( \bigcap \{Z_i = z_i\}\right)} = \frac{P\left(\bigcap \{Z_i = z_i\} \middle| \bigcap A_i \right)\left(\prod_i P\left( A_i\right)\right)}{\prod_i P\left( \{Z_i = z_i\}\right)},$$ but I don't how to proceed from here, or if it is even possible to do so.
EDIT: Maybe a better approach is to try showing that $\{P(A_i|Z_i)\}_i^n$ is an independent set (of random variables)? In this case, it seems trivial since $P(A_i|Z_i) = E(1_{A_i} | Z_i)$ is $\sigma(Z_i)$-mble and $\{\sigma(Z_i)\}_i$ are independent. Is this enough to show that independence holds in the above sense?
The answer is yes, as follows
$$P\left(\bigcap A_i \middle| \bigcap \{Z_i = z_i\} \right) = \frac{P\left(\bigcap A_i \cap \bigcap \{Z_i = z_i\} \right)}{P\left(\bigcap \{Z_i = z_i\} \right)} = \frac{P\left(\bigcap \{A_i \cap \{Z_i = z_i\}\} \right)}{P\left(\bigcap \{Z_i = z_i\} \right)} = \prod_i \frac{P\left(\{A_i \cap \{Z_i = z_i\}\} \right)}{P\left(\{Z_i = z_i\} \right)}= \prod P(A_i |Z_i = z_i)$$