I am guessing that the answer is "yes" if you interpret the question in the following way. Let $A_i$ be some subalgebras of a von Neumann algebra $(M,\tau)$ and assume that there are mutually orthogonal Hilbert subspaces $H_i$ of $H=L^2(M)$ so that for all $i$, $x (H\ominus H_i) \subset H_i$ whenever $x\in A_i$ with $\tau(x)=0$.

Let us also assume that $1 \perp \oplus_i H_i$ (probably this is not necessary).

Then if $y = x_1 \dots x_n$ with $x_j \in A_{i(j)}$, $i(1)\neq i(2)$, $i(2)\neq i(3)$, etc. and $\tau (x_j) = 0$, we have:
$x_n 1 \in H_{i(n)}$ since $1\in H\ominus H_i$;
$x_{n-1} x_n 1 \in H_{i(n-1)}$ since $x_n 1 \in H_{i(n)} \subset H\ominus H_{i(n-1)}$ (because $i(n)\neq i(n-1)$ and so $H(i(n))\perp H(i(n-1))$;
$x_{n-2} x_{n-1} x_n 1 \in H_{i(n-2)}$ since $x_{n-1} x_n 1 \in H_{i(n-1)}\subset H\ominus H_{i(n-2)}$, etc. Thus We get that $x_1\dots x_n 1 \in H_{i(1)} \perp 1$, so that $\tau(y)=0$. It follows that $A_1,\dots,A_n$ are freely independent.

(Conversely, if $M$ is generated by $A_1,\dots,A_n$ and they are free inside of $M$, then $L^2(M) = \mathbb{C}1 \oplus \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$, where $L^2_0(A_j) = \{1\}^\perp \cap L^2(A_j)$. Then you can take $H_j = \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots; j_1= j} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$ and then $H_j$ are orthogonal and
$H\ominus H_j$ is taken to $H_j$ by any $x\in A_j $ with $\tau(x)=0$).

If you now make some assumption (e.g. that $A_j$ are finite-dimensional, abelian or hyperfinite) then it follows from Ken Dykema's results (see e.g. his paper on Interpolated free group factors in Duke Math J.) that the von Neumann algebra they generate inside of $M$ is an interpolated free group factor. This is similar to the assumption you have put on the group (since the subgroup generated by a single element in the ping-pong lemma is necessarily abelian).

On the other hand, you raise the much bigger question of whether there exists some criterion that singles out free group factors -- just as the various functional-analytical criteria were shown by Connes to be equivalent to hyperfiniteness. Unfortunately, not much in known in this direction (note that a similar question exists on the ergodic equivalence side of things: is there a functional-analytic way of recognizing treeable actions? Or Bernoulli actions of free groups?)

any criterionat the purely vN level which characterizes the class of free group factors within the class of all group vN algebras -- but there are people reading MO who would no much more about this than I do. (Of course that class might consist of only one vN alg up to isomorphism, which is one reason I'm slightly pessimistic.) – Yemon Choi Jan 31 '11 at 18:25