Try $n=10^{k_1}+\cdots+10^{k_s}$ with distinct $k_i$, then $s(n)=s$. We have $n^2=\sum_1^s 10^{2k_i} +2\sum_{i<j} 10^{k_i+k_j}$ and we can choose the $k_i$ such all terms appearing have a different amount of factors 10. Hence $s(n^2)=s+s(s-1)=s^2$. Hence ${{s(n^2)}\over {s(n)}}=s$ and we can take $s=R$.