No, your strip does not surject onto $Y(2)$.
Domains $A \subseteq \mathbf{H}$ have a $\operatorname{PSL}(2,\mathbb{R})$-invariant "area" given by
$ \int_{A} \frac{\mathrm{d}x\, \mathrm{d}y}{y^2} $
Your strip has area 4, but the standard fundamental domain for $\operatorname{PSL}(2, \mathbb{Z})$ has area $\frac{\pi}{3}$. As $\Gamma(2)$ has index 6 in $\operatorname{PSL}(2,\mathbb{Z})$ and $6 \times \frac{\pi}{3} > 4$, there's no way it can surject onto $Y(2)$.
(PS: Here's another reason: $\Gamma(2)$ has 3 cusps, and your set contains a neighbourhood of $\infty$ but not of any other cusp. So there's no way that any set contained in $\{ z: \operatorname{Im}(z) > r\}$, for any $r > 0$, can surject onto $Y(2)$.)