Of course not. E.g., suppose that the $X_i$'s are independent random variables each uniformly distributed on the interval $(0,1)$. Let $f_m(X_i):=B_{i,m}-1/2$, where $B_{i,m}$ is the $m$th digit in the binary expansion of $X_i$. Then all your conditions hold, with $\sigma^2=1/4$. Also, the $f_m(X_i)$'s are i.i.d.
So, $$P\Big(\max_{1\le m\le M} \max_{1\le t\le T} \frac{S_{m,t}}{\sqrt{T}}>x\Big) =1-P\Big(\max_{1\le t\le T}\frac{S_{1,t}}{\sqrt{T}}\le x\Big)^M \\ \ge1-P\Big(\frac{S_{1,T}}{\sqrt{T}}\le x\Big)^M \underset{M\to\infty}\longrightarrow1 \not\le C x^{-2} \sigma^2$$ for any real $C>0$ and any $x>\sigma\sqrt C$ such that $$P\Big(\frac{S_{1,T}}{\sqrt{T}}\le x\Big)<1,$$ which will hold if $T>4x^2$. $\quad\Box$