I almost surely believe that it is not, but can't find a mistake in my argument below:
Fix a manifold $M$ and consider orientable plane bundles. These have $GL^+(2,\mathbb{R})$ structure group, which reduces to $SO(2)\cong U(1) \cong S^1$. So on the one hand orientable plane bundles are classified by $[M,BS^1]$ because $S^1$ is the structure group. On the other hand, $U(1)$ structure tells me that these are identified with complex line bundles which correspond to $H^1(M,\mathbb{Z})$ under the first chern class. But $H^1(M,\mathbb{Z})\cong [M, K(1,\mathbb{Z})]\cong [M,S^1]$.
So I have $[M,S^1]\cong [M,BS^1]$.
The first Chern class of a complex vector bundle is an element of degree two cohomology, not degree one. So the relevant identifications are $H^2(M) \cong [M, K(\mathbb{Z}, 2)] \cong [M, \mathbb{CP}^{\infty}]$.
Instead, we have $BS^1 = \mathbb{CP}^{\infty}$. In general, if $K(G, n)$ is a topological group (which is the case if $n > 1$, or $n = 1$ and $G$ is abelian), then it follows from the long exact sequence in homotopy applied to the universal $K(G, n)$-bundle $K(G, n) \to EK(G, n) \to BK(G, n)$ that $BK(G, n) = K(G, n+1)$.