Well, sure, but it's probably not the answer you're looking for: clearly $K$ fits the bill.

What you are probably looking for is the sub-ring generated by $R$ and $S$ (I think you can characterize this ring as $\{r+s+r's':r,r' \in R, s,s' \in S\}$). Since this ring is contained in $K$ it has no zero-divisors, and is thus an integral domain. Since its field of fractions is contained in any field that contains it, $\text{Fr}(\langle R\cup S\rangle) \subseteq K$.

Since this field of fractions also contains $R$ and $S$, it must also contain $\text{Fr}(R),\text{Fr}(S)$, and by the minimality of $K$, it must BE $K$.