Since R is an integral domain it has a well defined fraction field; call this field K. We know that K is a subfield of F. If F is any larger then F contains an x such that x and 1/x are not in R. Therefore F is precisely the fraction field of R.
The integers (in the rationals) don't make it, because R contains neither 3/2 nore 2/3. However, R could be all the fractions that do not have a factor of p in the denominator, for some prime p. Verify that R is a ring, and that it contains either x or 1/x. It doesn't contain 1/p, but it contains p, so we're ok.
An extension of a valuation ring, within its fraction field, is still a valuation ring, as it has either x or 1/x.