Generally Speaking, Cohen-Macaulay condition is an open condition in a moduli. So in general, even if a special fiber is not Cohen Macaulay, we should not expect that generic fiber is not Cohen-Macaulay.
But here is a special case: if the Groebner degeneration of an irreducible subscheme of a product of projective spaces is not Cohen-Macaulay, under what conditions can we say that the original irreducible subscheme of a product of projective spaces is also not Cohen-Macaulay?
I will answer in the contrapositive. Let $X \subseteq \prod_i {\mathbb P}^{n_i}$ be irreducible of codimension $k$. If whenever $\sum k_i = k$, you can find subspaces $\prod_i {\mathbb P}^{n_i}$ that intersect $X$ in at most one point, then $X$ is called a "multiplicity-free subvariety".
In this case Brion has proven that any degeneration of $X$ must still be Cohen-Macaulay: http://arxiv.org/abs/math/0211028 and in particular, $X$ itself must be!
The case of the diagonal was studied by Cartwright and Sturmfels: http://arxiv.org/abs/0901.0212
Just to add one comment (which may well follow from Allen's answer): if the degeneration is regular in codimension 1 yet has a point of codimension > 1 (i.e., the generic point of an irreducible closed subset of codimension > 1) at which it is both locally reduced and locally disconnected (i.e., removing the irreducible closed subset locally disconnects the degeneration), then the generic fiber should not be Cohen-Macaulay.
