No.
Such a bound would imply a similar bound on $$\displaystyle \left \lvert \sum_{N(z) = M} \left(\frac{z}{w} \right)_3 \right \rvert.$$
If $M$ is a product of distinct primes $p_1,\dots p_n$ congruent to $1$ mod $3$, the norms of primes $\pi_1,\dots,\pi_n$ then $$\sum_{N(z) = M} \left(\frac{z}{w} \right)_3 = \left( \left(\frac{1}{w} \right)_3 + \left(\frac{-1 }{w} \right)_3 \right) \left( \left(\frac{1}{w} \right)_3 + \left(\frac{\omega }{w} \right)_3 + \left(\frac{\omega^2 }{w} \right)_3 \right) \prod_{i=1}^n \left( \left(\frac{\pi_i}{w} \right)_3 + \left(\frac{\overline{\pi_i} }{w} \right)_3 \right)$$
so if we choose $w$ such that $\left(\frac{-1 }{w} \right)_3 =\left(\frac{\omega }{w} \right)_3 =1$ and then choose $n$ different primes $\pi$ such that $ \left(\frac{\pi_i}{w} \right)_3 =\left(\frac{\overline{\pi_i} }{w}\right)_3$ then this will have size $6 \cdot 2^n$.
Taking $n$ sufficiently large, we contradict any bound like the one you request.