I don't know a place where this is written up explicitly, but it's not hard as soon as one has an appropriate standard model for a neighborhood of a symplectic manifold for use in the Weinstein Neighborhood theorem.
A model that I've found useful in other contexts (see also p. 12 of this paper) is as follows. Let $X\subset Y$ be a symplectic submanifold of the symplectic manifold $(Y,\omega)$, and let $\pi: E\to X$ denote the symplectic normal bundle to $X$ in $Y$. $E$ is then naturally a symplectic vector bundle, and choosing a compatible fiberwise complex structure $I$ on $E$ induces a Hermitian metric $\langle\cdot,\cdot\rangle$ on $E$. Choose a unitary connection on $E$, inducing a horizontal-vertical splitting $TE=T^h E\oplus T^v E$, with respect to which we may write an arbitrary tangent vector $u\in T_eE$ as $u^h+u^v$ with $u^v$ naturally identified as an element of the fiber $E_{\pi(e)}$. Now define $L:E\to \mathbb{R}$ by $L(e)=\frac{\langle e,e\rangle}{4}$, and define a $1$-form $\theta\in \Omega^1(E)$ by $\theta(u)=dL(Iu^v)$ and then a $2$-from $\Omega$ on $E$ by $$ \Omega = \pi^*(\omega|_X) - d\theta $$
There is an obvious identification of $TE|_X$ with $TY|_X$, and under this identification the restrictions of $\Omega$ and $\omega$ are equal. Hence the Weinstein neighborhood theorem we get a symplectomorphism from a neighborhood of $X$ in $(Y,\omega)$ to a disk-bundle in $(E,\Omega)$. Moreover the Hamiltonian $L$ on $E$ generates a circle action which just rotates the $\mathbb{C}^k$ fibers of $E$ in the standard way. Performing the symplectic cut using this circle action replaces an appropriate-radius disk-normal bundle to $X$ with the quotient of the sphere normal bundle by the Hopf map (i.e. with the complex projectivization of the normal bundle), thus seeming to give what you want. Evidently this reduces to the standard construction of the blowup when $X$ is a point.
