Well, I don't see any way to get easy expressions. But for messy expressions, we have: $$E[Y] =\sum_{i=1}^n 2p_i(1-p_i) $$ Now suppose two strings differ by $y\geq 1$ bits. Enumerate these $\{1, \ldots, y\}$. For $k \in \{1, \ldots, y\}$ define $A_k=1$ if the first string has a 1 in the $k$th different bit, and $A_k=0$ else. Then $A_1, \ldots, A_y$ are i.i.d. Bernoulli with $Pr[A_k=1]=1/2$, and, given $Y=y$: $$ X|_{Y=y} = \left|\sum_{k=1}^y A_k - \left(y - \sum_{k=1}^y A_k\right)\right| = \left|y-2\sum_{k=1}^y A_k\right| $$ So: $$ E[X|Y=y] = \sum_{r=0}^y {y \choose r} (1/2)^y\left|y-2r\right|$$ Then: $$ E[X] = 0 + \sum_{y=1}^n \underbrace{Pr[Y=y]}_{complicated}E[X|Y=y]$$ Also: $$ E[XY] = E[YE[X|Y]] = 0 + \sum_{y=1}^{n} yPr[Y=y]E[X|Y=y] $$
Perhaps a more useful relationship is just the observation: $$ X = |Y-2R| $$ where $R = 0$ if $Y=0$ and $R$ is a sum of $Y$ iid Bernoulli $(1/2,1/2)$ variables else.