The claim in the title seems very plausible since the characteristic function "characterizes" or determines the distribution of $X$, but I don't know how to derive it. There is a similar result for the characteristic functions of two Random variables, eg here, but I'm not sure if it could be deduced from that.
Any help would be appreciated!
On
Reading my own post one year later I would supply the following answer. From Kac's theorem, to prove independence of $X $ and $1_F $ it would be sufficient to prove that for any $s,t \in \mathbb R^d $ we have that
$$E[e^{i \langle (X, 1_F), (s,t) \rangle}]= E[e^{i \langle X , s \rangle } ]E[e^{i \langle 1_F , t \rangle } ] $$
But since $e^{i \langle 1_F , t \rangle } = 1_{F^C } + 1_F e^{i\sum_{i=1} ^d t_i} $ and $e^{i \langle (X, 1_F), (s,t) \rangle}= 1_{F^C } e^{i \langle X, s \rangle} + 1_F e^{i \langle X, s \rangle}e^{i\sum_{i=1} ^d t_i} $ - from the fact that expectation is additive and we may "move out" constants proving the condition in the title of the question would be sufficient.
If $Z$ is any bounded function measurable with respect to $\mathcal F$ then $Z$ is a uniform limit of simple functions measurable with respect to $\mathcal F$. From this it follows that $Ee^{i\langle x, X \rangle } e^{i\langle y, Y \rangle } =Ee^{i\langle x, X \rangle } Ee^{i\langle y, Y \rangle}$ for all $x,y$ and for all $\mathcal F$ measurable random variable $Y$. It follows that $X$ is independent of $Y$ for all $\mathcal F$ measurable random variable $Y$.