This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.
Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).
Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.
