Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

1) The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

2) I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

3) So, let's see how one could construct a counterexample. Perhaps cones will do...

4) Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

5) So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

6) So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. I think this might give you what you want:
Grothendieck duality gives that then $$R\pi_*\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$
Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as
$$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1} $$
So, if we also knew that $R^i\pi_*\omega_Y=0$ for $i>0$, then it would follow that $R\pi_*\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition:

If $R^i\pi_*\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume.

Any thoughts?