This is an extended comment. 

To summarize, you have RVs $U\to X\to Y\to V$ where:

* $X\sim B(1/2)$,
* $Y=X+Z_Y$
* $U=X+(X\cdot Z_{U1} + (1-X)\cdot Z_{U0})$, 
* $V=Y+(Y\cdot Z_{V1} + (1-Y)\cdot Z_{V0})$, 

for RVs:

* $Z_Y\sim B(p_Y)$,
* $Z_{Ui}\sim B(p_{Ui})$, $i=0,1$
* $Z_{Vi}\sim B(p_{Vi})$, $i=0,1$

You are essentially wondering whether $U',V'$ with $U'\to X\to Y\to V'$ and:

* $U'=X+Z_{U'}$
* $V'=Y+Z_{V'}$

choosing: 

* $Z_{U'}\sim B(p_{U'})$ so that $I(U';X)=I(U;X)$
* $Z_{V'}\sim B(p_{V'})$ so that $I(V';Y)=I(V;Y)$

gives $I(U';V')\geq I(U;V)$. 

If all the $Z$'s turn out to be independent because of the Markov structure (not sure whether or not this is true), my inclination is that the answer to this question is **always no for nontrivial cases, i.e. when (U,X) or (Y,V)  are not BSCs**

This is because there is a notion that uniform noise is the most degrading. Choosing $(X,U'),(Y,V')$ to be BSCs will make the noise distribution $(V'-U')$ not only uniform, but *also uniform over each intermediate hop,* i.e. 'maximally bad.'