The first half of my answer didn't make much sense. I deleted it and expanded the second half.

For this range an upper bound sieve should suffice. The real quantity is smaller than the main term by a factor $\frac{\log x}{\log\log x}$, so loosing a constant does not matter. Hence we can replace the set of primes e.g. by all integers having only prime factors $\geq x^{0.01}$. To do so by Selbergs sieve you need asymptotics for the number of solutions of $a^2+b^2\equiv 0\pmod{n}$, $a,b,\leq x$, $q|ab$, where $n$ and $q$ satisfy $n<x$, $q<$x^{0.02}$, which looks doable.

For example, McDaniel (On the divisibility of an odd perfect number by the sixth power of a prime. Math. Comp. 25 (1971), 383–385) showed that an odd perfect number is either divisible by the sixth power of a prime, or it is not divisible by any prime $<100$. The proof is quite easy, once one has shown that if $n$ is not divisible by the sixth power of a prime, then $n$ is not divisible by some small prime, but getting there is pretty difficult.

There is a lot of literature proving the non-existence of odd perfect numbers satisfying certain additional restrictions. Practically all of them require some starting point of the form "One of these primes divides $n$" or "One of these primes does not divide $n$". Sometimes this starting point is easy to get (e.g. if you bound the number of distinct prime factors), but quite often getting the starting point is the most difficult part.

Cohen and Hagis proved that the largest prime factor of an odd perfect number is $>10^6$. Of course, the proof is indirect, so it begins with "Let $n$ be an odd perfect number with all prime divisors $\leq 10^6$...". Then a sequence of properties for $n$ are deduced, one of which reads "Lemma: $n$ is not divisible by 3, 5, 7, ..." . This Lemma was later taken out of context and cited as "Let $n$ be an odd perfect number. Then $n$ is not divisible by 3,5,7,...".