Let $E$ be the total space of the sphere bundle $S^k\to E\to M$, is it true that there exists a disk bundle $D^{k+1}\to N\to M$ such that $E=\partial N$? (where $D^{k+1}$ is the unit disk in $\mathbb R^n$)

If you don't specify what structure group you want the disc bundle $D^{k+1} \to N \to M$ to have, then it is always true: you just take the fibrewise cone on the original family. If you want to know whether smooth $S^k$bundles always bound smooth disc bundles, this is true iff $$O(k+1) \to \mathrm{Diff}(S^k)$$ is a homotopy equivalence (known as the Smale conjecture). This is known to be true if $k=0, 1$ (classical), $2$ (Smale) or $3$ (Hatcher). I don't think it is known in any other dimension, and is definitely false in general. In fact, it is false on $\pi_0$ in general, due to the existence of exotic spheres and hence also exotic selfdiffeomoprhisms of spheres. 


The question of whether or not a smooth sphere bundle fibrewise extends to a smooth disc bundle over a space $X$ boils down to whether or not the classifying map $$ X \to BDiff(S^n) $$ lifts up $$ BDiff(D^{n+1}) \to BDiff(S^n)$$ Where $Diff(D^{n+1})$ is the group of diffeomorphisms of the disc. The map $Diff(D^{n+1}) \to Diff(S^n)$ splits as a product: $$O_{n+1} \times PDiff(D^{n}) \to O_{n+1} \times Diff(D^n rel \partial)$$ where $PDiff(D^{n})$ is the group of pseudoisotopy diffeomorphisms of $D^{n}$. These are diffeomorphisms of $D^{n} \times [0,1]$ that are the identity on $(D^{n} \times \{0\}) \cup (S^{n1} \times [0,1])$. There is a fibrebundle: $$Diff(D^{n+1} rel \partial) \to PDiff(D^{n}) \to Diff(D^n rel \partial)$$ so basically this is asking whether or not this bundle has a section. I think it can't have a section, in particular the map $\pi_1 Diff(D^n rel \partial) \to \pi_0 Diff(D^{n+1} rel \partial)$ is epic by Cerf's Pseudoisotopy theorem. Okay, so this is now an answer. So this is saying that there are sphere bundles over $S^2$ which do not extend to smooth disc bundles over $S^2$. 


No, it seems that the "fibered cobordism group" is not even necessarily finitely generated, much less trivial: "Some fibred cobordism groups are not finitely generated", by L. Astey, 1988. EDIT (or, duh) As pointed out by @jc in his comment, the reference indicates that the answer is YES, not NO. 

