Here is a quick proof that any complete local Cohan-Macaulay ring of dimension $1$ and multiplicity $2$ is a *hypersurface*, so the answer to your last question is always yes.

Call such ring $R$ with maximal ideal $m$. Since $e(R)=2$ and $R$ is CM, there is a regular element $x\in m$ such that $$length(R/xR) = e(R)=2$$ (see Bruns-Herzog, chapter 4). 

EDIT: as Graham pointed out below, technically for such $x$ to exist one needs $R/m$ to be infinite. But one can enlarge the field without affecting the conclusion that $R$ is a hypersurface. 

Now the left hand side is $length(m/xR) +1$, so $length(m/xR)=1$, so $m/xR$ is one-generated. It follows that $m$ is $2$-generated. Thus $R=A/I$, $A$ is a regular local ring of dimension $2$. But since $R$ is CM, $I$ must be of pure height one, and since $A$ is regular, $I$ is principal. 

Note: in fact, the dimension one condition is *not* necessary, you can take a full length regular sequence and argue the same way (with just a little more work). So any CM singularity with multiplicity $2$ is a hypersurface. 

EDIT: Once we know that the local ring at the singular point is a hypersurface, Jason Starr and Qing Liu's nice comments show that there must exist a smooth surface containing $C$, thus completely answer the question.