Let $Z_t= e^{\int_0^t\phi dB_s-\frac12\int_0^t\phi^2ds}$ (the standard stochastic exponential).
I would like to prove that if there exists a $\mu>0$ s.t. $$\sup_{0\le s \le T}E[e^{\mu|\phi_s|}]<\infty$$ then $E[Z_t]=1$.
I'm sorry that I don't provide part of my proof, but it really does not come any idea on my mind.
Assuming that $\phi$ is deterministic, note that $\int_0^{1} \phi (s)dB_s$ is normal with mean $0$ and variance $\int_0^{1}\phi^{2}$. This implies that $Ee^{\int_0^{1} \phi (s)dB_s} =e^{\int_0^{1}\phi^{2} /2}$. Just multiply this by $e^{-\int_0^{1}\phi^{2} /2}$ to get $EZ_t=1$.
I have used the standard fact that if $Y\sim N(0,\sigma^{2})$ then $Ee^{Y}=e^{\sigma^{2} /2}$.