Here is the answer for finitely generated modules of rank one. Recall that the isomorphism classes of these modules form a group, the Picard group $Pic(R)$, with tensor product as multiplication.
Theorem (Traverso, Swan)
For a commutative ring $R$ the following are equivalent:
a) The reduced ring $R_{red}=R/Nil(R)$ is semi-normal
b) The natural group morphism $Pic(R)\stackrel {\cong}{\to} Pic(R[U])$ is an isomorphism ($U=$ indeterminate)
And what does it mean that $R$ is semi-normal?
It means that if $x,y\in R$ satisfy $y^2=x^3$, then there exists $s\in R$ with $x=s^2$ and $y=s^3$ .
(Geometrically: you can parametrize the cusp over $R$ ).
Admittedly this condition is a little strange, but at least it is easy to see that a normal ring $R$ (= integrally closed domain) is semi-normal :
Take $s=\frac {y}{x}\in Frac(R)$. Of course we have $x=s^2$ and $y=s^3$.
The key point is that $s\in R $ : the fraction $s=\frac {y}{x}$ is integral over $R$ because it satisfies the monic equation $T^2-x=0$ and since $R$ is integrally closed we must have $s\in R$ .