Note that $F$ depends on how you enumerate the rationals - for each enumration $e:\mathbb{N}\rightarrow\mathbb{Q}$, we get an $F_e$. And it is not obvious that the question does not depend on which $e$ you pick!
I'll show that there is some $e$ whose associated $F_e$ is not perfect. We'll enumerate the rationals in such a way that $\pi$ (for example) is in $F_e$ but is isolated. To do this, we pick a pair of monotonic sequences of rationals: $q_i\rightarrow \pi$ from below, and $p_i\rightarrow\pi$ from above. As long as these sequences converge to $\pi$ "sufficiently quickly" (say, with $p_i-q_i<2^{-{2^i}}$), we can cook up an $e$ such that the open balls $e$ puts around the $q_i$s union to form an open interval whose right endpoint is $\pi$, and the open balls $e$ puts around the $p_i$s union to form an open interval whose left endpoint is $\pi$; and furthermore, $\pi$ is never covered by any of the open balls on any of the points. This ensures that $\pi$ is an isolated point of $F_e$.
I believe, meanwhile, that there are enumerations of $\mathbb{Q}$ which do yield perfect sets, however. Why? Well, pick some enumeration $d$. The associated $F_d$ is an uncountable closed set, so we can write $F_d=P\sqcup C$ where $P$ is the maximal perfect subset of $F_d$ and $C$ is countable. Now, note that no point in $C$ can be a limit point of points in $P$, since the points in $C$ are exactly those points with Cantor-Bendixson rank $<\infty$. But that means we can find an open $U$ containing $C$ with $U\cap P=\emptyset$.
Indeed, we can find such a $U$ with arbitrarily small measure. So if we just had a little extra measure, we could smoosh things around and cover up $P$ - and this would yield a perfect $F$. But where can we get that extra measure?
Well, in any covering of $\mathbb{Q}$ by open balls, there's lots of overlapping. So it seems likely that we could tweak our enumeration $d$ to get a $d'$ which covers exactly $P^c$. This, however, would take some work to verify, and I'm not certain it's correct.
