The model you specified is, in fact, partial in the sense that you only have conditional expectation/variance of $Y$ given $X$. In order to answer your question, you need to know the value $E[X|Y]$. This is possible when you have marginal distribution of $X$. So as whuber pointed out in the comment, it depends on the marginal distribution of $X$. If your model is $Y=\beta X+U$, the answer is yes if and only if $E[X|Y] = \frac{1}{\beta}Y$
Some special cases when $U$ is normal. If $X$ is normal, then your answer is YES as you have independence. If $X$ follows a Bernoulli with $p=\frac{1}{2}$ and $\beta=1$, you can easily see that $E[X|Y] \neq Y$ is never true.