Let $X=TS^2\setminus Z$ where $Z$ is the zero section of the tangent bundle of $S^2$.


Is there a $S^2$- fiber bundle structure on $(X,\mathbb{R}^2\setminus\{0\},q)$ for some continuous fibre map $q$? 



Is $TS^2\setminus Z$ a $S^2$- fibre bundle on the puntured plane?(Swapping the role of fibre points and base space)
Assuming that your question means: Is $X$ a fiber bundle over $\mathbb{R}^2 -0$, the answer is no. $X$ is homotopy equivalent to the unit tangent bundle of $S^2$, which is in turn homotopy equivalent to $\mathbb{R}P^3$. Also, $R^2 -0 \simeq S^1$. Now look at the exact sequence of the fibration; you'd conclude $\pi\_2(\mathbb{R}P^3) = \mathbb{Z}$.


