Let $R=K[x_1,...,x_n]$ be the polynomial ring over a field $K$ and $I[x_1,...,x_n]=(u_1,...,u_t)$ be a Cohen-Macaulay monomial ideal of $R$. If $m<n$, could we say that $I[x_1,...,x_m,0,0,...,0]$ is Cohen-Macaulay in $K[x_1,...,x_m]$?
No, we can't say that. For a counterexample, take $n = 3$, $m = 2$, and let $I = (x_2^2, x_1x_2, x_1x_3)$.
To see that $I$ is Cohen-Macaulay, note that the Krull dimension of $R/I$ is 1. Cohen-Macaulayness is then equivalent to $I$ having no associated prime $P$ with dimension $R/P$ equal to 0. Since $I$ is homogeneous, its associated primes are homogeneous, and so the only possible example of such an associated prime is $P = (x_1,x_2,x_3)$, which is easily seen not to annihilate any element of $R/I$, hence is not associated with $I$.
On the other hand the restricted ideal $I' = (x_2^2, x_1x_2) \subset K[x_1,x_2]$ is not Cohen-Macaulay, since $x_2$ is annihilated by $(x_1,x_2)$ in $K[x_1,x_2]/I'$, and so $(x_1,x_2)$ is an associated prime of $I'$.