Is there any holomorphic version of the tubular neighborhood theorem?

I don't have a proof that this works, but according to Joe Harris, if $C \subset \mathbb{P}^2$ is a smooth conic curve, then there is no holomorphic map from a neighborhood of $C$ to $C$, something that would be doable if there was a holomorphic tubular neighborhood theorem. Additionally, he claimed that, if you let $X$ be the total space of the normal bundle of $C$, with $C \subset X$ as the zero section, then even the first order neighborhoods of $C$ in $X$ and in $\mathbb{P}^2$ are not isomorphic. Again, I lack proof, but the second statement is something that I assume can be checked algebraically.

Aside: there is another way that you could try to cook up a counterexample. If there was a tubular neighborhood theorem, with $X \subset Y$ complex manifolds, let $Z$ be the total space of the normal bundle of $X$ in $Y$. Then one has that $T_Z|_X = T_X \oplus N_{Y/X}$ in $Z$, and so if you follow through the supposed map in the holomorphic tubular neighborhood theorem, you get that $T_Y|_X = T_X \oplus N_{Y/X}$.

Edited to add: here is something that I can show is a counterexample. Let $\pi: X \rightarrow V$ be a non-isotrivial proper family of curves of genus greater than $2$. Choose a smooth fiber $C = \pi^{-1}(p)$. Then a neighborhood of $C$ contains $\pi^{-1}(U)$ for some neighborhood $U$ of $p$, and I have that $\pi^{-1}(q) \neq \pi^{-1}(p)$ for a dense open set of all $q \in U$. Then, any map from $\pi^{-1}(U) \rightarrow C$ must factor through $\pi$: any fiber not isomorphic to $C$ admits no nonconstant maps to $C$ and so any holomorphic map from $\pi^{-1}(U)$ factors through $\pi$ on a dense open and so it does so on all of $\pi^{-1}(U)$. But then, if the holomorphic tubular neighborhood theorem were true, there would be a holomorphic map from $\pi^{-1}(U) \rightarrow C$ that restricts to the identity on $C$, which is in contradiction with what was just remarked.