What does $\#P\subseteq FP^{PPAD}$ imply?
First, $\mathrm{PPAD\subseteq FP^{NP}}$, hence $\mathrm{\#P^{PPAD}\subseteq\#P^{NP}\subseteq FP^{\#P}}$. Moreover, $\mathrm{PPAD}$ is closed under Turing reductions, i.e., $\mathrm{FP^{PPAD}\subseteq PPAD}$. Thus, if we assume
$$\mathrm{\#P\subseteq PPAD},$$
then
$$\mathrm{\#P^{PPAD}\subseteq PPAD},$$
which by induction implies
$$\mathrm{FCH=PPAD}.$$
Passing to decision problems, since $\mathrm{P^{PPAD}\subseteq P^{TFNP}\subseteq NP\cap coNP}$, this shows
$$\mathrm{\#P\subseteq PPAD}\implies\mathrm{CH=P^{PPAD}=NP=coNP}.$$
(Note that using the closure of $\mathrm{PPAD}$ under Turing reductions, $\mathrm{P^{PPAD}}$ consists of predicates whose characteristic functions can be computed as projections of $\mathrm{PPAD}$ problems.)

As for $\oplus\mathrm P$, I believe $\mathrm{PPA\supseteq PPAD}$ can be solved by binary search on the predicate “the sum of degrees of vertices whose labels start with a given string is odd”, which means
$$\mathrm{PPAD\subseteq FP^{\oplus P}},$$
thus (using $\mathrm{P^{\oplus P}=\oplus P}$)
$$\mathrm{\#P\subseteq PPAD}\implies\mathrm{CH=P^{PPAD}=NP=coNP=\oplus P}.$$
A similar argument applies with $\mathrm{Mod}_p\mathrm P$ in place of $\oplus\mathrm P$ for any prime $p$.

I don’t know about UP.