Complexity class when reducing decision problem to function problem
In the case of PSPACE and FPSPACE, the answer is yes (since DEC reduces to FUN and DEC is PSPACE-hard, FUN is FPSPACE-hard). The reason is that in PSPACE search always reduces to decision: say you want to compute a function $f(x)$. If the problem of deciding whether the $i$-th bit of $f(x)$ is 1, given $i$ and $x$, is in PSPACE, then $f$ itself is in FPSPACE. Now, if $f$ is in FPSPACE, we can use your function FUN to compute it by using FUN to compute the decision problem "is the $i$-th bit of $f(x)$ 1?" for each $i$. One way to say this is that PSPACE is FPSPACE-hard.

However, there are complexity classes where the relationship between the function and decision versions is not so clear. For example:


It is not known if the search version of Group Isomorphism (find an isomorphism) reduces to the decision version (is there an isomorphism?), when the groups are input as multiplication tables.
$P^{NP[\log]} = P^{NP}_{tt}$, but $FP^{NP[\log]} = FP^{NP}_{tt}$ is not known to hold and would imply $NP=RP$ and $P=UP$ (see Selman, A. A taxonomy of complexity classes of functions. JCSS 48(2):357-381, 1994.) But if $P^{NP}_{tt}$ were $FP^{NP}_{tt}$-hard, we would automatically have equality of the function classes.


(For reference: $P^{NP[\log]}$ is $P$ with an $NP$-oracle, but the $P$ machine is only allowed to make $O(\log n)$ queries to the oracle. $P^{NP}_{tt}$ is the problems that are polytime truth-table reducible to $NP$; this is equivalent to nonadaptive polytime Turing reductions, i.e. all the queries to be made are decided in advance, then all the queries are made, then some computation is done without queries to find the result.)