Your problem is NP-complete by the modern presentation of Schaefer's dichotomy theorem.
You can also prove its NP-completeness by direct reduction of SAT to your problem. If a CNF formula has $n$ variables $x_1 \dots x_n$ introduce $n$ new variables $y_1 \dots y_n$ and add 2-CNF clauses that force each $y$ variable to have the opposite value of the corresponding $x$ value.
$$(x_1 \lor y_1)$$
$$(\lnot x_1 \lor \lnot y_1)$$
$$(x_2 \lor y_2)$$
$$(\lnot x_2 \lor \lnot y_2)$$
$$(x_n \lor y_n)$$
$$(\lnot x_n \lor \lnot y_n)$$
All solutions to the resulting formula will have half the variables with true values and half false.
For unaltered formulas, you don't need a special algorithm to find half-true half-false solutions. Take any CNF formula with an even number of variables and then add CNF clauses encoding a chain of adder circuits that sums the values of all the variables. Next, add a comparison circuit that is only satisfied if the sum is one-half the total number of variables. Run a normal SAT solver on the result and it will output only solutions where one-half the original variables have a true value. This is a somewhat naive solution but it gets the job done.
Search the literature a bit more and you will discover SAT solvers that accept pseudo-Boolean constraints in addition to CNF. With these you can write out the $x_1 + x_2 + x_3 + \dots + x_n = n / 2$ equation explicitly and the solver will either churn out the adder circuits for you or just handle the equation internally.
Your special SAT problem is also readily transformed into a 0-1 integer programming problem for which there are solvers available.