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.
To sketch a proof of a bound, let's first understand
$$\left \lvert \sum_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)_3 \right \rvert$$
where $\phi$ is a smooth bump function. The trivial bound is $O(RT)$ as we are summing over an annulus of radius $R$ and thickness $T$. Another bound is provided by integrating the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ against the Fourier transform of $\left(\frac{z}{w} \right)_3 $. Since the Fourier transform of $\left(\frac{z}{w} \right)_3 $ is a bunch of Gauss sums, the same at every point, the bound we get this way will be $\sqrt{N(w)}$ times the $L^1$ norm of the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $.
For $T$ small, the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ will look like the integral of a phase over a circle, i.e. a Bessel function, thus, viewed as a function of a complex variable $x$, approximately constant for $|x| < R^{-1}$ and proportial to $1/\sqrt{x}$ for larger values of $x$. The bump function will give us a rapidly decreasing cutoff at a radius of $1/T$.
Since the value at $0$ should be $RT$, the Fourier transform will take the value $\approx RT$ for $|x| <R^{-1}$ and $\approx R^{1/2} T / |x|^{1/2}$ for $ R^{-1} < |x| < T^{-1}$. The $L^1$ norm of this is $$ \int_{ R^{-1} }^{T^{-1} } (R^{1/2} T/ r^{1/2} ) \cdot 2 \pi r \cdot dr \approx R^{1/2} T (T^{-1})^{3/2} = (R/T)^{1/2} $$ so we get
$$\left \lvert \sum_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)_3 \right \rvert \ll \min ( RT, (R N(w) /T)^{1/2} )$$
Now in the sharp cutoff case we can express the characteristic function interval of radius $\sqrt{A}$ and thickness $B/ \sqrt{A}$ as a sum of smoothed characteristic functions of the same radius and dyadically decreasing thickness. So the bound we get will be obtained by summing the bounds for (dyadically) different values of $T$ from $0$ to $B/\sqrt{A}$, and thus will be comparable to the bound for the worst case of $T$.
The worst case occurs when the two terms in the minimum are equal, $ RT= (R N(w) /T)^{1/2}$, or $T = (N(w)/ R)^{1/3}$, giving a bound of $$ R^{2/3} N(w)^{1/3} = (A N(w))^{1/3},$$ in other words, by working out the details, I believe it should be possible to prove a bound of the form
$$\displaystyle \left \lvert \sum_{A < N(z) < A + B} \left(\frac{z}{w} \right)_3 \right \rvert \ll (A N(w))^{1/3}$$
This gives cancellation for $A \gg N(w)^{1/2}$, so we still obtain cancellation in intervals of greater than square-root size, which makes sense as this is possible in the classical setting (https://arxiv.org/abs/1508.00512), but not as much cancellation for larger intervals, which also makes sense as the boundaries in this setting are less smooth than in the classical one.
I am not sure exactly what Webb had in mind, but here is my answer. There is no real difference between the Selberg sieve in integers and in the polynomial ring $\mathbb{F}_q[T]$. So you can consult the book "Multiplicative Number Theory I" by Montgomery and Vaughan where the same upper bound is established in integers, and adapt their argument to $\mathbb{F}_q[T]$. Specifically see page 94, where a sum analogous to the one you ask about is estimated.
In more detail, the required estimate is
$$\sum_{\substack{D \in \mathscr{D} \\ \lvert D \rvert \leq N^{1/4} \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} \geq c \log^2(N).$$
For given $d\ge 0$ with $q^d \le N^{1/4}$, I claim that
$$(\star)\quad\sum_{\substack{D \in \mathscr{D} \\ \deg(D)=d \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} =q^{-d}\sum_{\substack{\deg(D)=d \\(D,K)=1}} 2^{\omega(D)} \mu^2(D) \ge c' d$$
if $d$ is sufficiently large (in terms of $K$). The constant $c'$ depends on $K$. Summing this over $d$ gives the needed bound.
To establish $(\star)$ write the arithmetic function $f(D)=2^{\omega(D)}\mu^2(D)\mathbf{1}_{(D,K)=1}$ as a convolution of the divisor function $\tau(D)$ with a simple multiplicative $g$. Since $\sum_{\deg(D)=d} \tau(D) = q^ d (d+1)$, the problem reduces to showing $\sum_{D}g(D)/\lvert D \rvert$ converges to a positive constant and $\sum_{D} g(D) \deg(D)/\lvert D \rvert$ converges absolutely. This is easily verified by considering the Dirichlet series of $g$ (as in page 94 of the aforementioned book).
Best Answer
Based on the ideas posted by @OfirGorodetsky and @tomos as comments to this post, I managed to compile this solution.
$$\begin{align*} S &= \sum_{m = M}^{2M} \frac{\phi(m)}{m} \prod_{p \nmid md}\left(1 - \frac{1}{p^2} - \frac{\chi_p(m)}{p^2} \right) \; \chi_d(m) \\ &= \sum_{m = M}^{2M} \frac{\phi(m)}{m} \left( \sum_{r \nmid md} \prod_{p \nmid mdr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi_r(m)}{r^2} \right) \chi_d(m) \\ &= \sum_{m = M}^{2M} \prod_{p | m} \frac{1}{1 + \frac{1}{p}} \sum_{r \nmid d} \prod_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi_{rd}(m)}{r^2} \\ &= \sum_{r \nmid d} \left( \mu(r) \frac{1}{r^2} \prod_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \right) \sum_{m = M}^{2M} \left( \prod_{p | m} \frac{p}{1 + p} \right)\chi_{dr}(m) \end{align*} $$
Now setting $f(m) = \prod_{p | m} \frac{p}{1 + p}$ and $g = f*1$, by the Mobius inversion formula we have that $g(m) = \sum_{ab = m} f(a) \mu(b)$. It is easy to see by computation that $|g(m)| < \frac{1}{m}$. Hence we have,
$$\begin{align*} \left| \sum_{m = M}^{2M} \left( \prod_{p | m} \frac{p}{1 + p} \right)\chi_{dr}(m)\right| &= \left| \sum_{m = M}^{2M} \sum_{ab = m} g(a)\chi_{dr}(a) \chi_{dr}(b) \right| \\ &= \left| \sum_{a = 1}^{2M} g(a)\chi_{dr}(a) \sum_{b = M/a}^{2M/a} \chi_{dr}(b) \right| \\ &\leq \sum_{a = 1}^{2M} |g(a)| \left| \sum_{b = M/a}^{2M/a} \chi_{dr}(b) \right| \\ &\ll \sum_{a = 1}^{2M} \frac{1}{a} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{dr} \log{(dr)} \end{align*}$$
So now, $$\begin{align*} |S| \ll \sum_{r \nmid d} \frac{1}{r^2} \log{(M)} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{d} \log{(d)} \end{align*}$$
The Euler factor involving the $\rho_m$ function was too complicated to get a useful function applying Mobius inversion. So somehow had to expand out and dealt as shown.