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.
If $M$ is projective, then there is module $K$ s.t. $M \oplus K \cong R^{(\Lambda)}$. For any multiplicative set $S \subset R$, one gets $S^{-1}M \oplus S^{-1} K \cong (S^{-1}R)^{(\Lambda)}$. Hence, any localization of a projective module is projective.
If $(R,m)$ is local and $M$ is finitely generated projective then $M$ is free. Here is a proof. The short exact sequence
$$
0 \to K \to R^n \to M \to 0,
$$
splits, hence $K$ is finitely generated and sequence
$$
0 \to K \otimes R/m \to R^n \otimes R/m \to M \otimes R/m \to 0,
$$
also splits. By Nakayama's lemma we can choose generators of $M$ in such way that the map $R^n \otimes R/m \to M \otimes R/m$ is an isomorphism, then $K \otimes R/m \cong K/mK=0$. By Nakayma's lemma $K=0$, since it is finitely generated.
Fact: if $M$ is finitely presented, then $S^{-1}Hom_R(M,N) \cong Hom_{S^{-1}R} (S^{-1}M, S^{-1}N)$.
To check that $M$ is projective is enough to check that for any surjection $N \to N'$ map $Hom(M,N) \to Hom(M, N')$ is surjective, but this can be checked stalkwise if $M$ is finitely presented (by the fact).
There are plenty of other proofs, but often they involve flat modules, in this proof we only use two ways to characterize projective modules and the fact. Notice that this proof doesn't use that $R$ is Noetherian.
Best Answer
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.