Is there an Oka-Grauert principle for homogeneous spaces? Suppose we have a fibration over the punctured disc (i.e., a deformation of complex manifolds) such that each fiber is a homogeneous space. Is the total space a product of a fiber with the punctured disc? What about the case where the fiber is Fano?
 A: To elaborate on my comment: yes, there is an Oka-Grauert principle for homogeneous spaces. The definite reference (besides the earlier papers of Grauert and Gromov) is the book "Stein manifolds and holomorphic mappings" by Forstneric.
In there, you find Corollary 5.4.8 telling you that for a holomorphic fiber bundle  $\pi:Z\to X$ over a reduced Stein space $X$ with Oka manifold fibers, the inclusion of holomorphic into continuous sections is a weak equivalence.  Proposition 5.5.1 tells you that every complex homogeneous manifold is an Oka manifold (via the exponential spray from the corresponding complex Lie group). Combining these two, any continuous section of a holomorphic fiber bundle with homogeneous space fibers can be deformed to a holomorphic section. Therefore, a holomorphic fiber bundle over the punctured disc with homogeneous space fibers is holomorphically trivial if it is topologically trivial. 
Now that we have translated the problem into algebraic topology, we can get rather complete information. The fiber bundle with fiber $G/H$ over the punctured disc $\Delta^\ast$ is completely classified by the homotopy class of the monodromy map $\pi_1(\Delta^\ast)\to\pi_0(\operatorname{Aut}(G/H))$, i.e., by a connected component of the holomorphic automorphisms. Therefore, any homogeneous space with non-connected automorphism group can be used to manufacture non-trivial families. In abx's answer, the quadric $Q\cong\mathbb{P}^1\times\mathbb{P}^1$ has an automorphism - the map which switches the two factors - which is not homotopic to the identity, giving rise to a topologically non-trivial family. On the other hand, there are two cases where the bundle is topologically (and by the above also holomorphically) trivial: 1) $\operatorname{Aut}(G/H)$ is connected, or 2) the fiber bundle with fiber $G/H$ is the associated bundle for a principal $G$-bundle, and $G$ is connected.
A: The relevant paper is "Local rigidity of quasi-regular varieties" by Pasquier and Perrin,
Math. Z. 265 (2010), no. 3, 589–600. They construct a smooth fibration over $\mathbb{C}$ such that the fiber over $t\neq 0$ is a orthogonal grassmannian $\mathbb{G}_q(2,7)$, but the fiber over 0 is not homogeneous.
Edit: Actually I overestimated the question. As it stands, just take the family of quadrics given by  $X^2+Y^2+Z^2+tT^2=0$ in $\mathbb{P}^3\times \mathbb{C}$. The family cannot be trivial over the punctured disk because the monodromy exchanges the two generators of $H^2$ of a smooth fiber
