So is it possible to use modular square root Oracle for factorizing Integers in a far more efficient way than using the number field sieve ?
Assuming your modular square root Oracle is far more efficient than NFS, yes.
Here's one straight-forward approach (where $N$ is the number of be factored):
Pick a random number $r$ uniformly from the random $[1, N-1]$
Compute $r^2 \bmod N$ and give it to your square root Oracle
That Oracle will return with a number $s$ with $s^2 = r^2 \pmod N$
Compute $t = \gcd(r+s, N)$
Assuming $N$ is an odd semiprime, then with probability 0.5, $t$ will be one of the prime factors of $N$ (and the other prime factor can easily derived)
You can see that all of the above steps (other than possibly the Oracle itself) can be efficiently computed. And, if you repeat this until you get a factorization, you get an expected number of Oracle queries of 2.
Why this algorithm works is actually fairly easy to see; for an odd semiprime, then for any nonzero quadratic residue, there are four square roots. Because $r^2 = s^2 \bmod N$, this means that $(r-s)(r+s) = kN$, for some integer $k$. That means that of the two factors $p, q$ of $N$, each could be either in $r-s$ or in $r+s$, and all four possibilities happen in the four possible square roots. Because the square root is essentially random (because $r$ was chosen at random and independently of the Oracle), we have probability 0.5 that $p$ and $q$ will end up in different places, and hence $r+s$ will have one of $p, q$ as a factor, and not the other (and hence the $\gcd$ will recover it)
