Your ${\cal F}$ is actually ${\cal G}$, that is the already enlarged filtration/probability space. So, the claim here seems to be that we do not have to consider the smaller, market filtration, ${\cal F}$.
But, before we invoke Hypothesis (H), only this is true:
$$ E\left[1_{\tau>T}|{\cal G}_t\right] = 1_{\tau>t} E\left[e^{\Gamma_t -\Gamma_T}|{\cal F}_t\right]\left(\not= e^{\Gamma_t -\Gamma_T}\right), $$
when $\gamma_t$ is a stochastic process (also note the presence of $1_{\tau>t}$ even when it is deterministic).
More generally, for $X$ ${\cal F}_T$-measurable (and integrable), we have:
$$ E\left[X1_{\tau>T}|{\cal G}_t\right] = 1_{\tau>t} E\left[Xe^{\Gamma_t -\Gamma_T}|{\cal F}_t\right]\left(\not= E\left[X|{\cal F}_t\right]e^{\Gamma_t -\Gamma_T}\right). $$
Hypothesis (H) is equivalent to $\sigma$-algebras $ {\cal F}_\infty $ and ${\cal G}_t$ being conditionally independent given ${\cal F}_t$ under $Q$, for all $t\geq 0$. It is also equivalent to:
$$ P(\tau \leq t |{\cal F}_t)=P(\tau \leq t |{\cal F}_\infty) $$
for all $t\geq 0$.
As it happens, the canonical construction of default time, based on a given ${\cal F}$-progressively measurable hazard rate process $\gamma_t$ (does not need to be deterministic) and a random variable $\zeta$ which is uniform on $[0,1]$ and independent of ${\cal F}$, supported by the enlarged space $(\Omega, {\cal G}, P)$, with $$ \tau := \inf \; \{t\geq 0| e^{-\Gamma_t} < \zeta \}, $$ automatically implies that Hypothesis (H) holds. (Cox processes do too.)
Indeed:
$$ P(\tau >t |{\cal F}_\infty)= P(e^{-\Gamma_t} \geq \zeta |{\cal F}_\infty) = e^{-\Gamma_t}$$
and
$$ P(\tau >t |{\cal F}_t)= E[P(e^{-\Gamma_t} \geq \zeta |{\cal F}_\infty)|{\cal F}_t] = e^{-\Gamma_t},$$
as $\Gamma_t$ is ${\cal F}_t$-measurable.
More on Hypothesis (H) information interpretation here.