Let $D$ be an integral domain which is not a field and $Q=\text{Frac}(D)$ the field of fractions of $D$. Then $Q$ as a $D$-module has not a projective cover.
By Corollary 5.35 of Rotman's Homological Algebra we got that : If $D$ is an integral domain and $Q=\text{Frac}(D)$, then $Q$ is a flat $D$-module.
So we got it, $Q$ is a flat $D$-module, now my idea is to use Bass Theorem to kill this one, this theorem states (among other things) that $M_{R}$ has projective cover ($M_{R}$ is perfect) iff every $M_{R}$ flat module is projective.
So I reduced (or complicated :S) the problem to prove that the flat module $Q_{D}$ is not projective, which is the part I cannot prove. I tried to study the proof of $\mathbb{Q}_{\mathbb{Z}}$ is a flat module that is not projective but they use the fact $\mathbb{Q}_{\mathbb{Z}}$ is finitely generated and that $\mathbb{Z}$ is DIP but in my case I dont know if $Q_{D}$ is finitely generated, also $D$ is not be DIP. Any help in order to prove this problem in the direction I propose or any other will be apreciated. Thanks!
As a hint, note that $Q=\text{Frac}(D)$ is also injective as an $R$-module. Therefore to show that it is not projective you can show that there are no non-zero projective-injective modules over $D$.
There is also a solution to this problem on this site, which can be found here. This is actually exercise 3.18 in Rotman's Intro to Homological Algebra, 2nd edition, so you might want to try and prove it using his hints.