Here are some sufficient conditions:
Let $A$ be a local domain, $K$ its fraction field, $L$ a finite extension of $K$, and $B$ the integral closure of $A$ in $L$. If $A$ is Noetherian and integrally closed, and $L$ is separable over $K$, then $B$ is necessarily finite over $A$, and so is semi-local (by going-up/down-type theorems).
If $A$ is not integrally closed, but its integral closure in $K$ is finite over itself, then again $B$ will be finite over $A$.
The condition that the integral closure of $A$ in $K$ be finite over $A$ is (at least by some people) called N-1. More generally, the condition that $B$ be finite over $A$ is called N-2, or Japanese. (The letter N here stands for Nagata, and I believe Grothendieck coined the adjecive Japanese for these rings because these properties were studied by Nagata and the commutative algebra school around him in Japan.)
So if $A$ is a Japanese ring, then $B$ will be finite over $A$, and hence semilocal. Of course, this is rather tautological: its utility follows from the fact that many rings (indeed, in some sense, most rings --- i.e. most of the rings that come up in algebraic number theory and algebraic geometry) are Japanese. E.g. all finitely generated algebras over a field, or over $\mathbb Z$, or over a complete local ring, are Japanese.
Here are some useful wikipedia entries related to this topic: Nagata rings and Excellent rings.
The fraction field $K$ has the structure of an $R$-module, and for any maximal ideal $\mathfrak{m}$ of $R$, it also naturally has the structure of an $R_\mathfrak{m}$-module. If $K$ is finitely generated as an $R$-module, then $K$ must also be finitely generated as an $R_\mathfrak{m}$-module because $R_\mathfrak{m}\supseteq R$. Thus we are reduced to the case when $R$ is local.
Let $R$ be a local domain with maximal ideal $\mathfrak{m} \neq 0$. Clearly, $\mathfrak{m}K=K$. But the Jacobson radical of $R$ is just $\mathfrak{m}$, so if $K$ is finitely generated as an $R$-module, then Nakayama's lemma implies that $K=0$, which is a contradiction.
Best Answer
If the fractions $p_i/q$ ($i=1,2,\dots,n$) form a set of generators (it's not restrictive to assume the denominators are the same), then any element of $F$ can be written as $$ \frac{x}{y}=\sum_{1\le i\le n}\frac{p_i}{q}d_i =\frac{1}{q}\sum_{1\le i\le n}p_id_i $$ which means $1/q$ is a generator. Since $F\ne D$, we have $1/q\notin D$. Then $$ \frac{1}{q^2}=\frac{d}{q} $$ for some $d\in D$, which means $dq=1$, a contradiction.
If $1\notin D$, we have that $\frac{q}{q^2}$ is a generator, but then $$ \frac{q}{q^3}=\frac{qd}{q^2} $$ and so $$ \frac{q}{q^2}=q\frac{q}{q^3}=q\frac{qd}{q^2}=\frac{q^2d}{q^2}=d $$ which is again a contradiction.