Does the martingale property holds after changing filtration?

Question

Let $\Omega$ be the space of continuous real-valued functions $\omega=(\omega_t)_{t\ge 0}$ starting at zero, i.e. $\omega_0=0$. Let $\Lambda=\Omega\times \mathbb R_+$ and denote by $\lambda=(\omega,\theta)$ its elements. Define by $\left(B=(B_t)_{t\ge 0},T\right)$ the canonical elements, i.e. $B(\lambda)=\omega$ and $T(\lambda)=\theta$. Define further the canonical filtration $\mathbb F=(\mathcal F_t)_{t\ge 0}$, i.e. $$\mathcal F_t~:=~\sigma\left(B_s, \{T\le s\} \mbox{ for all } s\le t\right).$$ It is clear that $T$ is a $\mathbb F-$stopping time. Now let $\mathbb P$ be a probability measure on $(\Lambda, \mathcal F_{\infty})$ s.t. $B$ is a $\mathbb F-$Brownian motion under $\mathbb P$. Now, let $X$ be a gaussian random variable and $\varphi: \mathbb R\times\mathbb R\to\mathbb R$ be a bounded measurable function. Denote by $\mathbb F^B=(\mathcal F^B_t)_{t\ge 0}$ the natural filtration generated by $B$, then it is clear that the process $Z_t:=E\left[\varphi(B_T,X)~|~\mathcal F^B_t\right]$ is a continuous martingale w.r.t. $\mathbb F^B$. My question is whether $Z=(Z_t)_{t\ge 0}$ is also a martingale w.r.t. to $\mathbb F$? Many thanks for the reply!