This is true assuming that $Z_t$ is independent of the Markov chain. Indeed, then
$$d(t):=d_G(t)=\max_x E\|P_x(Z_t)-\pi\|_{TV},$$
where
$$P_x(n):=\delta_x P^n$$
and $\delta_x$ is the row matrix $([a_y]_{y\in\mathcal X})^T$ with $a_y:=1_{y=x}$. It is easy to see that for any probability measures $\mu$ and $\nu$
$$\|\mu-\nu\|_{TV}=\sup_{0\le f\le1} \int f\,d(\mu-\nu), \tag{0}$$
where $\sup_{0\le f\le1}$ is taken over all measurable functions $f$ such that $0\le f\le1$.

Take now any real $s$ and $t$ such that $1\le s\le t$. We have to show that then $d(t)\le d(s)$. So, it suffices to show that for each $x$
$$E\|P_x(Z_t)-\pi\|_{TV}\overset{\text{(?)}}\le E\|P_x(Z_s)-\pi\|_{TV}. \tag{1}$$
The random variable $Z_t$ is stochastically greater than $Z_s$. So, without loss of generality, $Z_t\ge Z_s$. Take now any column matrix $f=[f_x]_{x\in\mathcal X}$ with $f_x\in[0,1]$ for all $x$. Then the entries of the (random) column matrix $P^{Z_t-Z_s}f$ are in the interval $[0,1]$ as well and hence
$$(P_x(Z_t)-\pi)f=(P_x(Z_s)-\pi)P^{Z_t-Z_s}f\le\|P_x(Z_s)-\pi\|_{TV}$$
by (0).
So, again by (0),
$$\|P_x(Z_t)-\pi\|_{TV}\le\|P_x(Z_s)-\pi\|_{TV}.$$
Taking now the expectations, we get (1), as desired.