The conjecture is most commonly stated with rational coefficients: if $X$ is a simply-connected finite CW complex with an almost-free action of $T^n$, then 
$$
\operatorname{rk} H^\ast(X;\mathbb{Q})\ge 2^n.
$$
In the rational case the almost-free and free cases are equivalent. The reason is that if $X$ admits an almost-free $T^n$ action then it is rationally homotopy equivalent to a finite complex $X'$ with a free $T^n$ action, and of course $H^\ast(X;\mathbb{Q})\cong H^\ast(X';\mathbb{Q})$. This is originally due to Halperin; you'll find a nice discussion in the book *Algebraic Models in Geometry* by Félix, Oprea and Tanré, Section 7.3 (in particular Proposition 7.17).