Here is the work I've done so far. I believe my hang-up is that I'm still learning the basic properties of quotient rings and not so much because of anything about tensor products. I want to say I can factor out the "scalar" bits and be left with something that works out nice but I"m not sure cause commutativity isn't a given for $Q_R/R$ right?
$\mathbf{Theorem}$: For $R$ an integral domain with $1$ and $Q_R$ its field of fractions show that $(Q_R/R) \otimes_R (Q_R/R) = 0$.
$\mathbf{Proof}$: Let $r, s, e, d \in R$ and $d^{-1}, e^{-1} \in Q_R$.
Elements of $Q_R/R$ are the cosets $rd^{-1} + R$.
Elements of $(Q_R/R) \bigotimes_R (Q_R/R) $ are the cosets $(rd^{-1} + R) \otimes (se^{-1} + R)$
Since R has a unit and $se^{-1} + R$ = $dd^{-1}se^{-1} + R$
$(rd^{-1} + R) \otimes (se^{-1} + R) = (rd^{-1}d + R) \otimes (d^{-1}se^{-1} + R) $
$= (r + R) \otimes (d^{-1}se^{-1} + R) = (0 + R) \otimes (d^{-1}se^{-1} + R) = 0$
And since so the tensor product in question is the trivial $R$-module. $\square$
This is not something you need to over think. Given a basic tensor $p\otimes q$, there exists a 'integer' (element of $R$) such that $rp\in R$. This basic tensor is equivalent to $pr\otimes q/r=0\otimes 0$, and that is it.