Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R \subseteq M \subsetneq K$) is projective over $R$, then is $R$ integrally closed in $K$ ?
If this is not true in general, what if we strengthen the hypothesis to say every proper $R$-submodule of $K$ is projective over $R$ ?
This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $r\in R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $r\in R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $a\in P$ be nonzero and let $M=a^{-1}P\subset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $M\to P$, and thus $P$ is free. Since $P\subset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.