The symplectic blow-up of a compact symplectic manifold $(X,\omega)$ along a compact symplectically embedded submanifold $(M,\sigma)$ results in another compact manifold $(\tilde{X},\tilde{\omega})$ given by $$\tilde{X}=\overline{X-V}\cup_{\varphi} \tilde{V}$$ where $V$ is a tubular neighborhood of $M$ that is diffeomorphic via $\varphi$ to $\tilde{V}$ and lives away from the zero section in the canonical line bundle over the projectivization of the normal bundle of $M$ in $X$. The blow-down map $f:\tilde{X}\rightarrow X$ restricts to a diffeomorphism in $\overline{X-V}$, so in general, if my understanding is correct, $\tilde{X}$ may not be realized as a fibre bundle over $X$. Indeed, the cohomology algebra for $\tilde{X}$, as shown by McDuff, is a direct sum of the cohomology for $X$ with a finitely generated module over the cohomology of $M$. The Leray-Hirsch theorem gives a tensor product result for the cohomology of fibre bundles and product spaces.
My question is, is it known what circumstances precipitate the symplectic blow-up $\tilde{X}$ to be at least homotopically equivalent to a fibre bundle over $X$? Or is this nonsense?
This really isn't even true when $M$ is a point (this is really the only case I am familiar with). The symplectic blow-up is the natural generalization of the blowup in the complex setting (i.e. the birational isomorphism $\Bbb CP^2 \# \overline {\Bbb CP} ^2 \to \Bbb CP^2$). This is relatively clearly not a fiber bundle, and it would be completely impossible the blowdown to be a fiber bundle projection. In fact, say if $X$ is a symplectic 4-manifold, any symplectic blowup of $X$ at a point is diffeomorphic to $ X \# \overline {\Bbb CP}^2$.
If you really want, you can even pick compatible almost complex structures such that the blowup is biholomorphic to the (almost) complex blowup and the projection maps commute. In the big McDuff-Salamon book, there is a proof of Gromov non-squeezing using this construction, which can probably be read without reading much of the rest of the book.