In fact, for any finite extension $K$ of rationals (not just cubic), and for any non-$0$ ideal $J$ (not just $J=(d_K)$) it is true that any ideal class contains an ideal prime to $J$. If $J=\prod P_i^{a_i}$ and $A=C\prod P_i^{b_i}$ where $C$ is prime to $J$ then take $\alpha=\beta/\gamma$, $\alpha\in K$, $\beta,\gamma\in O_K$, such that $(\gamma)=D\prod P_i^{b_i}$ ($D$ prime to $J$) and $D|(\beta)$, $J$ prime to $(\beta)$ (such $\beta,\gamma$ exist by Chinese remainder theorem; even better, use a strong approximation theorem to get $\alpha$ directly). Then $B=\alpha A\subset O_K$ is prime to $J$.