I couldn't help thinking that there must be a lower-power proof of this. Here's what I came up with; I'd be interested if there are shorter proofs that are as elementary.

First, as both of the previous answers pointed out, unique divisibility makes $Q$ torsion-free; if there were non-zero $x\in Q$ and $n\in\mathbb N$ with $nx=0$ then division of 0 by $n$ wouldn't be unique.

Consider elements $a\in A$ that are divisible in $A$, meaning that $nx=a$ has solutions $x\in A$ for all non-zero integers $n$. I claim that each coset $C$ of $F$ contains a divisible element. Proof by contradiction: Suppose that each $c\in C$ failed to be divisible by some non-zero integer $n_c$. Let $m=\prod_{c\in C}n_c$, and use the assumption that the quotient $Q$ is divisible to get some $a\in A$ with $ma\in C$. But then this $ma$ is an element $c\in C$ divisible (in $A$) by $m$ and thus by $n_c$, a contradiction.

Next, I claim that each coset $C$ of $F$ contains only one divisible element. To see this, suppose there were two, say $c$ and $c'$. Then $c-c'$ is an element of $F$ that is divisible in $A$. In particular, $c-c'=|F|x$ for some $x\in A$. (Here $|F|$ denotes the order of $F$.) Projecting to $Q$ and using the fact that $c$ and $c'$ were in the same coset $C$, we get $0=|F|\bar x$ in $Q$, where $\bar x$ is the projection of $x$ into $Q$. But $Q$ is torsion-free, so $\bar x=0$, which means $x\in F$. But then $|F|x=0$, so $c=c'$.

So every coset of $F$ in $A$ contains a unique divisible element. The sum and the difference of two divisible elements are obviously divisible. So the divisible elements of $A$ constitute a subgroup of $A$ complementary to $F$, and we have the desired splitting.