As I commented above, I think the answer is that a $\psi$-space need not be an FZ-space, and that a counterexample may be constructed from a Luzin gap. Here are the details, which did not fit into the comment.
We first construct the MAD family, which will give the counterexample. Start by splitting ${\mathbb N}$ into two infinite sets $X_0,X_1$. Then let ${\mathcal A}_0=\{A^0_\alpha:\alpha<\omega_1\}$ be an arbitrary AD family on $X_0$ and ${\mathcal A_1}=\{A^1_\alpha:\alpha<\omega_1\}$ a Luzin gap on $X_1$ (i.e. an AD family which, in particular, has the property that no uncountable pair of subfamilies can be separated; here ${\mathcal B},{\mathcal C}\subseteq{\mathcal A}_1$ can be separated if there is a set $S\subseteq X_1$ which almost-contains all sets from $B$ and is almost disjoint from all sets in $C$.).
We shall fuse these two families into a single new AD family which we will extend to a MAD family. Proceed as follows: for $\alpha<\omega_1$ let
$A_\alpha=A^0_\alpha\cup A^1_{\alpha+1}$ and let $\{B_\alpha:\alpha<\omega_1\}$ be the enumeration of $\{A^1_\beta:\beta\in Lim(\omega_1)\}$. Let ${\mathcal A}$ be an arbitrary MAD family extending the family
$\{A_\alpha,B_\alpha:\alpha<\omega_1\}$.
Proposition The space $\psi({\mathcal A})$ is not an FZ-space.
Note In the following we identify the points of $\psi({\mathcal A})={\mathbb N}\cup D$ which are in $D$ with the elements of the MAD family ${\mathcal A}$.
Proof.
Define $f:\psi({\mathcal A})\to{\mathbb R}$ as follows: $f(n)=1/n$ for $n\in X_0$
and $f(x)=0$ otherwise. It is clear that $f$ is continuous. Let $G=\psi(\mathcal{A})\setminus f^{-1}(0) = X_0$. To show that $\psi({\mathcal A})$ is not an FZ-space it will be sufficient to show that $F=cl(G)$ is not a zero set. Notice first that for each $\alpha<\omega_1$ we have $A_\alpha\in F$ and $B_\alpha\not\in F$. Suppose now that $g:\psi({\mathcal A})\to{\mathbb R}$ is a continuous function such that $g^{-1}(0)=F$. Since $B_\alpha\not\in F$
for each $\alpha<\omega_1$, we can fix $n<\omega$ such that ${\mathcal B}=\{B_\alpha:g(B_\alpha)>1/n\}\subseteq{\mathcal A}_1$ is uncountable. Now let $S=\{k<\omega:g(k)>1/n\}$. Then $S$ almost contains all sets from ${\mathcal B}$ while, by continuity of $g$, being almost disjoint from every set in $\{A_\alpha:\alpha<\omega_1\}$ hence also from every set in ${\mathcal C}=\{A^1_{\alpha+1}:\alpha<\omega_1\}\subseteq{\mathcal A}_1$. This contradicts the fact that ${\mathcal A}_1$ is a Luzin gap, since $S$ separates the two uncountable subfamilies ${\mathcal B,C}\subseteq{\mathcal A}_1$. $\square$
