The short answer is no, since if we take the quadratic twist of $A$ by $\sqrt{D}$, then the conductor of $A$ will more-or-less acquire divisibility by the primes dividing $D$. So we can make the conductor arbitrarily large without changing the CM field $K$. For example, if $D$ is a square-free odd integer, then the conductor of $E_D:y^2=x^3+Dx$ is divisible by $D^2$, but the CM field $\mathbb{Q}(i)$ is independent of $D$.

On the other hand, you might find it useful to know that for any prime $\mathfrak{p}$, if $A$ has potential good reduction at $\mathfrak{p}$ (which is the case for CM abelian varieties), then the power of $\mathfrak{p}$ dividing $N_A$ is bounded by a constant that depends only on $g$ and $[K:\mathbb{Q}]$ (independent of $A$ and $\mathfrak{p}$). For example, for an elliptic curve $E/\mathbb{Q}$ with potential good reduction at $p$, we have $\operatorname{ord_p(N_E)}\le 2$ for $p\ge5$, $\operatorname{ord_p(N_E)}\le 5$ for $p=3$, and $\operatorname{ord_p(N_E)}\le 8$ for $p=2$.