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. Note that they **all** immerse in $\mathbb R^4$, which is better than what the immersion conjecture would say $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$. 

One argument for this is to immerse the surface in $\mathbb R^3$, and take the normal bundle. You then modify the normal bundle in $\mathbb R^4$ to achieve any line-bundle type.  Line bundles over surfaces are basically just mod-2 $1$-cocycles, so you define them by choosing curve systems where the normal orientation flips -- vaguely speaking.