The answer is $9$. Choose a curve $C$ with the required topology. Choose $8$ points $x_1$, $x_2$, ..., $x_8$ on $C_1$. The conditions of passing through the $x_i$ impose $8$ linear conditions on the $10$ dimensional space of cubics, so we can find a second cubic $E$ passing through the $x_i$ and not proportional to $C$. Let $x_9$ be the ninth intersection of $C$ and $E$.

I claim that $x_9$ is also on $C_1$. Proof: Consider $C_1$ as a loop in $\mathbb{RP}^2$, and let $\tilde{C}_1$ be the preimage in the universal cover $S^2$. The loop $C_1$ is not contractible so, if we travel all the way around $C_1$, we will go from $(x,y,z)$ to $(-x,-y,-z)$ in $S^2$. Since $E$ is an odd degree polynomial, $E(x,y,z) = - E(-x,-y,-z)$. So $E$ must change sign an odd number of times along $C_1$. We already know that it changes sign at $x_1$, ..., $x_8$; the only place to get an additional sign change is at $x_9$. (Note that this argument shows that $C_1 \cap D$ is always odd, counting with multiplicity. I think that a careful count will show that $C_1 \cap D_1$ is odd, and the other three possibilities are always even.)

Now, $E$ may not have the right topology. But, for $\epsilon$ sufficiently small, the curve $D= C+\epsilon E$ also passes through $x_1$, ..., $x_9$, and it does have the right topology.