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We have $T<\infty$ a.s. for T stopping time. Let $X_t$ be a continuous Martingale.

Is this enough to conclude $E[X_T]=E[X_0]$?

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No: let $X_t$ be a standard Brownian motion (starting at zero), and define $$ T=\inf\{t\geq 0:X_t=1\}$$ Then $T<\infty$ almost surely, hence $X_T=1$ almost surely, so $$ \mathbb{E}[X_T]=1\neq 0=\mathbb{E}[X_0] $$

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    Couldn't it be easier? A symmetric random walk also serves as a counter example, if I recall correctly. – 2017-01-15
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    @O. Von Seckendorff: Yes, but the question is about continuous-time martingales. – 2017-01-15
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    Thanks. I forgot Brownian motion was the "continuous random walk". – 2017-01-15
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    So, regarding the following theorem, i assume that the brownian motion is not uniformly integrable: Let $X$ be an adapted, càdlàg process with terminal value $X_\infty$. Then $X$ is an uniformly integrable martingale iff for all stopping times $T$ $$E[X_T]=E[X_0]$$ and $X_T$ is integrable holds? – 2017-01-15
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    @peer: I don't think I've ever seen uniform integrability characterized that way, but I wouldn't be surprised if that were true. – 2017-01-16
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    @carmichael561 You can look it up in Jacod/Shyraev: Limit theorems for Stochastic Processes. – 2017-01-26