There is a more refined version of Dirichlet's theorem which is immensely useful in these types of problems. It's called the Bombieri-Vinogradov theorem, which says roughly that the error term in Dirichlet's theorem cannot be very large for many values of $q$ simultaneously. While we can't bound any individual error term by $O(n^{1/2+\epsilon})$ (such a bound would be equivalent to a form of RH), we can get this level of control if we average over many values of $q$.
More precisely, let $E(q)$ denote the error $\left|\pi_{q^2,1}(n) - \frac{\pi(n)}{q(q-1)}\right|$ in Dirichlet's theorem. Then a simple consequence of Bombieri-Vinogradov is that for some constant $B > 0$, $\sum_{q < n^{1/4} \ \log^{-B} n} E(q) \ll \frac{n}{\log^2n}.$
Therefore the error terms for $q \le n^{1/4}\ \log^{-B} n$ cannot accumulate to exceed the $\Omega(n/\log n)$ difference between $\pi(n)$ and $\sum\limits_{q\text{ prime}} \frac{\pi(n)}{q(q-1)}$, exactly as you had hoped. The only moduli remaining are $n^{1/4} \ \log^{-B} n \le q \le \sqrt{n}$, but these are very easily controlled using the trivial bound $\pi_{q^2,1}(n) \le \frac{n}{q^2}$ (there aren't many numbers congruent to $1\!\!\pmod{q^2}$ when $q$ is large).
A slightly more general application of Bombieri-Vinogradov should be able to produce the asymptotic result cited in @GerryMyerson's answer.
EDIT — Had the wrong upper bound for $q$, since the modulus is $q^2$, not $q$.
EDIT 2012/06/13: The asymptotic is probably not as immediate as I thought; it would require some subtlety with regard to the number of primes $q$ we can sieve out.