Balanced Max-2-SAT NP-Hardness
I think it is possible to reduce Max Cut to this problem: given a graph $G(V,E)$, make a variable for each vertex, and for each edge $(u,v)\in E$ make the two clauses $(x_u\lor x_v)$ and $(\neg x_u\lor\neg x_v)$. In the resulting instance, $x_u$ appears $d(x_u)$ times positive and the same number of times negative. Since Max Cut is still NP-hard on regular graphs, this proves that the version for which $k$ is the same for all variables is also hard.