The answer is "no" because in general there does not even exist a neighbourhood $U$ of your submanifold that holomorphically retracts to the submanifold.
In fact Finnur Lárusson has [proved] 1 the following  (warning: I have kept his notations, so as to faithfully quote him. In particular $[X]$ is $\mathcal O(X)$, the line bundle on $M$ associated to the divisor $X$)

Theorem Let $X$ be a connected hypersurface in a compact complex manifold M of dimension at least $2$. If
(1) $H^0 (M, TM) = 0$
(2) $dim H^0(M; [X]) \geq 2$, and
(3) $H^1(M; [X]^{-1} \otimes TM) = 0$,
then no neighbourhood of $X$ retracts holomorphically onto $X$.

Note that the hypotheses are very easy to satisfy: (1) holds for $K3$ surfaces for example and (2), (3) will follow from $X$ being sufficiently ample.

Here is a freely downloadable version of Lárusson's article, from his homepage.

The answer is "no" because in general there does not even exist a neighbourhood $U$ of your submanifold that holomorphically retracts to the submanifold. In fact Finnur Lárusson has proved the following (warning: I have kept his notations, so as to faithfully quote him. In particular $[X]$ is $\mathcal O(X)$, the line bundle on $M$ associated to the divisor $X$)

Theorem. Let $X$ be a connected hypersurface in a compact complex manifold $M$ of dimension at least $2$. If

1. $H^0 (M, TM) = 0$
2. $\dim H^0(M, [X]) \geq 2$, and
3. $H^1(M, [X]^{\vee} \otimes TM) = 0$,

then no neighbourhood of $X$ retracts holomorphically onto $X$.

Note that the hypotheses are very easy to satisfy: (1) holds for $K3$ surfaces for example and (2), (3) will follow from $X$ being sufficiently ample.

Here is a freely downloadable version of Lárusson's article, from his homepage.