Let $F_{1}\to E_{1}\overset{\pi_{1}}{\to}\mathcal{M}$ and $F_{2}\to E_{2}\overset{\pi_{2}}{\to}\mathcal{M}$ be two fibre bundles over the same base manifold $\mathcal{M}$. If $H:E_{1}\to E_{2}$ is a bundle morphism such that its restriction $H:E_{1p}\to E_{2p}$ to the fibres $E_{1,2p}:=\pi_{1,2}^{-1}(\{p\})$ is for all $p\in\mathcal{M}$ a diffeomorphism, I have to show that $H$ is a diffeomorphism and hence a bundle isomorphism.
But isn't that obvious? I mean $H$ is by assumption bijective, because $E_{i}$ is the disjoint union of all its fibres and $H$ is a bijective on all of them. Furthermore, $H$ is by definition smooth and since it is a diffeomomorphism for ach fibre, its inverse is smooth restricted to all fibres and hence smooth as a whole, since the total spaces are the disjoint union of all their fibres.
Edit: This answer addresses the case in which $H$ is a morphism of topological fiber bundles (i.e. merely continuous). The case where $H$ is required to be smooth is different, but the counterexample shows that the argument in the question is incorrect.
As Thorgott points out in a comment, the claim is false; an extra requirement is needed so that the diffeomorphism of fibers above $p$ "varies smoothly" as we vary $p$.
As a more concrete counterexample, take the bundle $\mathbb{R}^2\to\mathbb{R}$ given by $(x,y)\mapsto x$ and consider the map $\varphi(x,y)=(x,y+f(x))$. This map restricts to a diffeomorphism on each fiber (a translation), even if we chose a function $f$ which is not differentiable.