No, since for all $n \ge 2, S_{n} \equiv \{4,5\} \mod 6$, where $S_{n}$ and $S_{n+1}$ are not congruent mod 6. Since 5 is not a quadratic residue modulo 6, we are done. Edit: Wrong, $S_{10} \equiv 3 \mod 6$..

There are squares in the sequence $S_n$: $S_9=100$, $S_{2474}=5063^2$,...

Let $S_n=x^2$ and $S_{n+1}=y^2$. Let $p$ be the n+1-th prime. We have than that $(y-x)(y+x)=y^2-x^2=p$ since p is a prime from this follows that $y=x+1$ and $2x+1=p$. So we now have that $\frac{p-1}{2}^2$ should be the sum of the primes smaller than p. Notice that the sum of the numbers from $1$ to $p-2$ ($p-1$ is not prime for $p \neq 3$ ) is at most $x^2 \leq \frac{(p-2)(p-1)}{2} < \frac{p-1}{2}^2 = x^2$. Hence there is no solution. Rapid check of small cases and see that $S_3$ doesn't work.