For every $t$, let $U_t=(Z_t)^2$. Almost surely, the integral which defines $Y$ has infinite upper Riemann sums and zero lower Riemann sums. To see this, note that, on every interval of the subdivision, infinitely many independent random variables $(U_t)$ are involved, whose distributions have common support $S=[0,+\infty)$, hence their supremum is the supremum of $S$ and their infimum is the infimum of $S$.
As a consequence, almost surely the function $t \mapsto U_t$ is not Riemann(-Darboux) integrable and $Y$ does not exist. (That is, unless one precises another way to define the integral.)
Edit (Added to answer a question raised by @Sasha.)
As explained above, almost surely and for every partition $\{t_k\}_k$ of $[0,1]$, the upper Riemann sum is infinite and the lower Riemann sum is zero, that is, $ \sum\limits_k(t_k-t_{k-1})\sup\{U_s\mid s\in[t_{k-1},t_k]\}=+\infty, $ and $ \sum\limits_k(t_k-t_{k-1})\inf\{U_s\mid s\in[t_{k-1},t_k]\}=0. $ However, the Riemann sums $Y(P)$ associated to tagged partitions $P$ do converge uniformly to $1$ in $L^2$ when the mesh of $P$ goes to zero.
To see this, for any tagged partition $P=\{(s_k,t_k)\}_k$ of $[0,1]$, write $Y(P)$ for the Riemann sum associated to $P$. That is, $s_k$ is in $[t_{k-1},t_k]$ for every $k$ and $ Y(P)=\sum\limits_k(t_k-t_{k-1})U_{s_k}. $ Then, $E(U_s)=1$ and $\text{Var}(U_s)=3$ for every $s$, hence $E(Y(P))=1$ and, since the random variables $(U_{s_k})$ are independent, $ \text{Var}(Y(P))=3\sum\limits_k(t_k-t_{k-1})^2\le3\text{mesh}(P). $ This proves that $Y(P)\to1$ in $L^2$ when $\text{mesh}(P)\to0$, in the sense that $ \sup\{\|Y(P)-1\|_2\mid\text{mesh}(P)\le m\}\to0 $ when $m\to0$. In particular, for every sequence $(P_n)$ of tagged partitions, $Y(P_n)\to1$ (in $L^2$ and in particular, in probability) as soon as $\text{mesh}(P_n)\to0$.