Approximation of fiber bundle isomorphisms
This is true. Here is a high-powered proof. There are proofs in the flavor of Hirsch; I don't know a reference, but with some gusto you could prove them.
$C^k$ bundles with fiber $M$ are the same as fiber bundles with structure group $\text{Diff}^k(M)$. They are thus classified by maps $X \to B\text{Diff}^k(M)$, as long as $X$ is paracompact. Then your question follows by proving that the map $B\text{Diff}^\infty(M) \to B\text{Diff}^1(M)$ is a homotopy equivalence; because taking the loop space of this we recover the inclusion homomorphism, it suffices to show that $\text{Diff}^\infty(M) \to \text{Diff}^1(M)$ is a homotopy equivalence. But this follows from what's written in Hirsch:
Let $f: S^k \to \text{Diff}^1(M)$ be a sphere's worth of $C^1$ diffeomorphisms of $M$. By extending the smooth approximation theory in Hirsch to the case when the codomain is a Banach manifold, we may homotope $f$ to be $C^1$; this means that the induced map $f: S^k \times M \to M$ is $C^1$, not just $C^1$ in the $M$-direction. Now this is homotopic to a smooth map $f': S^k \times M \to M$; because $f(x,-)$ was a diffeomorphism and $f'$ is chosen sufficiently close to $f$, $f'(x,-)$ is also a diffeomorphism. (Diffeomorphisms are open in the $C^1$ topology.) Thus the map $\text{Diff}^\infty(M) \to \text{Diff}^1(M)$ is surjective on $\pi_k$. A similar argument in the case with boundary shows that it is injective on homotopy groups. Because $\text{Diff}^k(M)$ are metrizable manifolds, a theorem of Palais says that a weak homotopy equivalence between them is actually a homotopy equivalence, as desired.