There is an obvious homotopy (given by the usual inverted pant diagram) between the identity map from disc to disc and the map that deforms the separate holes to their wedge sum. But this does not in general imply that the images of the two maps are homotopy equivalent spaces, does it?
My reason for doubting that they are homotopy equivalent is the following: The disc with two holes is a compact surface with three boundary components, while the disc minus their wedge sum is a compact surface with two boundary components. And while I am unaware of the general theory, it seems to me that the number of boundary components is a homotopy invariant, at least within the class of compact surfaces.
No, indeed the two spaces have different fundamental groups, the free group on two generators for the first, respectively $\Bbb Z$ for a disk minus the wedge of two disks. (The first space retracts on a wedge of two circles, the second retracts on $\partial D$).
And indeed you are right, this number of component is also invariant, you can use the Alexander Duality as an alternative approach, but this is less elementary than simply computing fundamental group.