This is not true in this form.

Think about it this way: For simplicity let us assume that $D$ is already basepoint-free (and to give a counter-example this is certainly enough), so there exists a proper surjective birational morphism $f:X\to Z$ and a very ample line bundle $\mathscr L$ on $Z$ such that $\mathscr N:=\mathscr O_X(D)=f^*\mathscr L$. We may assume that $Z$ is normal and $f_*\mathscr O_X\simeq \mathscr O_Z$. Then for any $m\in\mathbb Z$,
$$
\tag{$\star$}
H^0(X, \mathscr N^{\otimes m})\simeq H^0(Z, \mathscr L^{\otimes m}),
$$
so in particular $Y\simeq \rm{Spec} (\oplus_m H^0(Z, \mathscr L^{\otimes m}))$ and since $\mathscr L$ is very ample, this means that $Y$ is a cone over $Z$.

Then you can connect the vanishing of the cohomology groups $H^i(Z, \mathscr L^{\otimes m}))$ to depth conditions on $Z$. In other words, it follows almost as you desire that
$$
H^i(Z, \mathscr L^{\otimes m})=0
$$
for all $m\in \mathbb Z$ and $0<i<\dim Z=\dim X$. However, this is exactly where the problem comes from.

While we do have $(\star)$ for $i=0$, in general we do not have that for arbitrary $i$'s. Instead what we have is a Leray spectral sequence, which essentially says that

$$
%\tag{$\star$}
H^i(X, \mathscr N^{\otimes m})\simeq H^i(Z, \mathscr L^{\otimes m}\otimes Rf_*\mathscr O_X).
$$

So, as long as $\mathscr O_X$ have higher direct images, you're out of luck.

However, this tells you how to fix the statement. You need to assume that $Z$ has rational singularities. Based on the above you should be able to put together a proof of that.

Finally, here is an explicit simple example when your desired statement fails: Let $Z\subseteq \mathbb P^3$ be a normal projective surface with a non-rational singularity. For example a cone over an elliptic curve. Let $f:X\to Z$ be a resolution of singularities. Choose a very very ample line bundle $\mathscr L$ on Z such that it has no higher cohomology and let $D$ be the divisor corresponding to $f^*\mathscr L$. Clearly this is semi-ample and big. Since $Z$ is a hypersurface, the cone over $Z$ (corresponding to the very ample line bundle $\mathscr L$) is Cohen-Macaulay, so your assumption on $Y$ is satisfied. Now let $\mathscr N:=\mathscr O_X(D)=f^*\mathscr L$ and consider the exact sequence provided by the obvious Leray spectral sequence:
$$
0 \to H^1(Z, \mathscr L) \to H^1(X, \mathscr N) \to H^0(Z, \mathscr L \otimes R^1f_*\mathscr O_X) \to H^2(Z, \mathscr L) \to \dots
$$
By assumption $H^1(Z, \mathscr L) =H^2(Z, \mathscr L) =0$, so
$H^1(X, \mathscr N) \simeq H^0(Z, \mathscr L \otimes R^1f_*\mathscr O_X)$, but the latter is strictly non-zero by the assumption that $Z$ has a non-rational singularity (which is necessarily isolated).