I'm going to assume your singularity is dimension $\geq 3$. Angelo beat me to the answer but he is right, this is not true. But it is true sometimes (including the Cohen-Macaulay case as he implied).
A singularity is normal if it is $R1$ and $S2$. In your case, an isolated singularity is normal if the depth at the singular point is at least 2.
Now, a general hyperplane section will be $R1$ by Bertini. So we just need to check that the general hyperplane is $S2$. Well, for this we just need the depth to be at least 2 again, and hence we just need the original singularity to have depth $\geq 3$.
Conclusion: If your singularity is $S3$ (in your case just $\text{depth} \geq 3$), then what you want holds after cutting down by ONE hyperplane
EDIT: As Angelo pointed out, the actual question didn't cut down by just one hyperplane. In that case you can't just have depth $\geq 3$, you need $X$ to be Cohen-Macaulay.
Of course, not all singularities satisfy this, for example a cone over an Abelian surface.
You might also look at this preprint which seems to have some related results: Tadashi Ochiai, Kazuma Shimomoto
