No, even in the most favorable case $(X_i)_{i\geqslant 0}$ iid with $\mathbb P(X_i=1)=\mathbb P(X_i=-1)=1/2$. Denoting $F_n$ the cumulative distribution function of $n^{-1/2}S_n$, we have by symmetry 
$$F_{2n}(0)=\frac 12(1+\mathbb P(S_{2n}=0)).$$
Since $\mathbb P(S_{2n}=0)=\binom{2n}n2^{-2n}$, denoting $\Phi$ the cdf of the standard normal distribution, 
$$\sup_{x\in\mathbb R}|F_{2n}(x)-\Phi(x)|\geqslant |F_{2n}(0)-\Phi(0)|\geqslant \frac 12\cdot \binom{2n}n2^{-2n}.$$
Using Stirling's formula, we obtain that the RHS behave asymptotically like $n^{-1/2}$.