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.