My vague memory is that we proved with Auinger 10 years ago that a freely indexed pro-supersolvable group is virtually pro-p. We were only interested in the case of relatively free groups, which appears in http://link.springer.com/article/10.1007%2Fs00208-006-0767-2

This case admits a number of simplifications and the published version is very different from the first proof we had, which I believe worked for what you want. I can't remember the details so let me outline a special case.

Suppose first that $G$ is finitely generated (and not pro-cyclic), freely indexed and pro-supersolvable with order divisible by only finitely many primes. Then by an old result of Oltikar and Ribes http://projecteuclid.org/euclid.pjm/1102806646 the Frattini subgroup of $G$ is open. The Frattini subgroup is also pro-nilpotent. An open normal subgroup of a freely indexed group is again freely indexed. A theorem of Lubotzky says that a pro-nilpotent freely indexed (and not procyclic) group is free pro-p for some prime p. Thus a freely indexed pro-supersolvable group with order divisible by only finitely many primes has Frattini subgroup open and free pro-p for some prime p.

So one wants to show in the general case that a freely indexed prosolvable groups has only finitely many prime divisors of its order.

We gave a geometric consequence for the Cayley graph of a freely indexed profinite group. Namely, any closed connected (in the sense of Ribes and Guldenhuys) subgraph of the Cayley graph contains each edge between two of its vertices. Using this, we essentially showed that such a profinite group is not a subdirect product of two profinite groups. This is how we showed a relatively free pro-solvable group which is freely indexed has only finitely many prime divisors. I am not sure if one can use this in general. I'd have to reread the paper more carefully or dig up some ancient versions of the paper on long lost computers to see if we really did do the general case at one point.

My vague memory is that we proved with Auinger 10 years ago that a freely indexed pro-supersolvable group is virtually pro-p. We were only interested in the case of relatively free groups, which appears in http://link.springer.com/article/10.1007%2Fs00208-006-0767-2

This case admits a number of simplifications and the published version is very different from the first proof we had, which I believe worked for what you want. I can't remember the details so let me outline a special case.

Suppose first that $G$ is finitely generated (and not pro-cyclic), freely indexed and pro-supersolvable with order divisible by only finitely many primes. Then by an old result of Oltikar and Ribes http://projecteuclid.org/euclid.pjm/1102806646 the Frattini subgroup of $G$ is open. The Frattini subgroup is also pro-nilpotent. An open normal subgroup of a freely indexed group is again freely indexed. A theorem of Lubotzky says that a pro-nilpotent freely indexed (and not procyclic) group is free pro-p for some prime p. Thus a freely indexed pro-supersolvable group with order divisible by only finitely many primes has Frattini subgroup open and free pro-p for some prime p.

So one wants to show in the general case that a freely indexed pro-supersolvable groups has only finitely many prime divisors of its order.

We gave a geometric consequence for the Cayley graph of a freely indexed profinite group. Namely, any closed connected (in the sense of Ribes and Guldenhuys) subgraph of the Cayley graph contains each edge between two of its vertices. Using this, we essentially showed that such a profinite group is not a subdirect product of two profinite groups. This is how we showed a relatively free pro-supersolvable group which is freely indexed has only finitely many prime divisors. I am not sure if one can use this in general. I'd have to reread the paper more carefully or dig up some ancient versions of the paper on long lost computers to see if we really did do the general case at one point.