David's answer reminds me also of the following:

For ordinary double points $R = k[[x,y,z]]/f$, $\text{A}_n$, $\text{D}_n$, $\text{E}_6$, $\text{E}_7$, $\text{E}_8$. These all have normalization in some field extension as a regular ring (they are all quotient singularities). Let $R \subseteq S$ be the usual extension with $S$ regular that $R$ is a quotient of. Then $S$ has exactly 1 $R$-summand, and so $S$ can't be free (and thus can't be flat either).

I know one way to to deduce this from a paper of Huneke-Leuschke, but I don't know the proof off the top of my head that there is clearly exactly one summand. Maybe Graham or Long does?

Finally, I should also add:

**Theorem:** Let $S$ be a module finite local extension of a regular local ring $R$ (for example, $S$ is the normalization of $R$ in some extension field of $K(R)$), then $S$ is Cohen-Macaulay if and only if $S$ is free = flat as an $R$-module.

Hm, maybe I'll also point out...

**Conjecture (Direct summand):** Let $R$ be regular, and $S$ the integral closure of $R$ in some finite extension of $K(R)$. Then $S$ has at least one $R$-summand (in other words, $R \to S$ splits as a map of $R$-modules). In particular, for any $R$-module $M$, $M \otimes R \to M \otimes S$ is injective.

This conjecture is known for rings containing a field and for rings in mixed characteristic of dimension $\leq 3$.