The answer is no. 

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle.  This is a considerably better number than $2n-a(n)$. 

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds. 

Consider for example the line bundles over surfaces. The immersion conjecture would say they all immerse in $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$ but I think we can argue they immerse in $\mathbb R^4$.  

Said another way, we are asking what line bundles are sub-bundles of normal bundles of immersed surfaces in $\mathbb R^4$.   For orientable surfaces you can quickly generate any isomorphism type.  For non-orientable surfaces I believe you can, as well.  You immerse the surface in $\mathbb R^3$ and use the $1$-cocycle as a way to guide how you "flip" the bundle, when thought of as inside the larger $\mathbb R^4$.