Relatively prime polynomials over integrally closed domain are(?) relatively prime over the fraction field

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I know that the following holds:

Lemma. Let $R$ be an integrally closed domain and $K$ its field of fractions. Let $f \in R[X] \setminus R$ monic. Suppose that there exist $g,h \in K[X] \setminus K$ such that $f=gh$. Then $g,h \in R[X]$. In particular, if $f$ is irreducible over R, then it's irreducible over K.

I have to prove the following:

Let $R$ be an integrally closed domain and $f,g \in R[X]$ relatively prime (i.e. there is no $h \in R[X] \setminus R$ such that $h|f$ and $h|g$). Then there are $p,q \in R[X]$ with $pf+qg \in R \setminus \{0\}$.

Question: is it true that if $f,g$ (not necessarily monic) are relatively prime in $R$, then they are relatively prime in $K$?

If this is true, then I can solve my exercise, since by Bezout's identity I can find $p',q' \in K[X] \setminus K$ such that $p'f+q'g = 1$, and then it's sufficient to multiply $p'f+q'g$ by the product of all the denominators of the coefficients of $p'$ and $q'$.