Your argument starts on the right track but you haven't said enough. A smooth $S^{k-1}$-bundle on a space $X$ is classified by a map $X \to B \text{Diff}(S^{k-1})$. The natural action of $O(k)$ on $S^{k-1}$ induces a map $BO(k) \to B \text{Diff}(S^{k-1})$, and the question is whether a map into $B \text{Diff}(S^{k-1})$ always admits a lift (up to homotopy) to a map into $BO(k)$.
The reason you haven't said enough is that this can be possible without the map $BO(k) \to B \text{Diff}(S^{k-1})$ being an equivalence. The lifting problem is always solvable iff it's solvable for the universal example, namely the identity map $B \text{Diff}(S^{k-1}) \to B \text{Diff}(S^{k-1})$. A lift of this map into $BO(k)$ is precisely a homotopy section of the natural map $BO(k) \to B \text{Diff}(S^{k-1})$ (that is, a section, up to homotopy). In particular, you don't need a homotopy inverse (as I think is being claimed in the comments), only a homotopy right inverse.
If a map between spaces has a homotopy right inverse then applying any homotopy-invariant functor produces a map with a right inverse, hence in particular a map which is surjective. So to rule this possibility out it suffices to show that the induced map on, say, $H_n$ for some $k$ fails to be surjective, or that the induced map on $H^n$ for some $k$ fails to be injective. But I don't know enough about $B \text{Diff}(S^{k-1})$ to do this.