There is no "vectorlike" gauge theory in the standard model, and this is a consequence of naturalness. This means that all particles in the standard model are naturally massless, and the mass only comes from Higgs mechanism. This is one of the great features of the standard model that is easy to break in any modification or extension.
The teminology "vectorlike" comes from the 1950s, when people didn't like 2-component spinors and thought that the world is fundamentally parity invariant. A "vectorlike" gauge field couples to a 4-spinor according to $\gamma^\mu A_\mu$, while a "pseudovectorlike gauge field" couples to a 4-spinor according to $\gamma^5\gamma^\mu A_\mu$. Both are parity invariant, but in the first case, A is a vector (meaning it changes sign under reflection), and in the second case it's a pseudovector.
But the gauge fields in nature are neither vectors or pseudovectors, they are parity-violating. They couple as "V-A" meaning $(1-\gamma_5)\gamma^\mu A_\mu$, which is a projection operator to one two component part of the 4 component Dirac spinor. This means that 4-component language is a little obfuscatory for this (although 4 component spinor notation is still useful, becuase Feynman trace identities are easier than Fierz identities, and the 4-component notation most easily generalizes to higher dimensions). The point is that there is no parity, and the gauge fields are neither "vectors" or "pseudovectors", they are parity violating vector fields which don't have a definite transformation under parity, because nature is chiral.
So I would drop the "vectorlike" terminology, and use the term "naturally mass allowing". A vectorlike gauge theory is "naturally mass allowing" because you can make the Fermion massive. This means that the left and right partners have the same charges, and this can be considered an accident.
The correct question is "why are all gauge theories in nature mass forbidding?" This is true of all the fields on the standard model--- none of the right handed and left handed fields in the standard model can pair up to form a mass, because they are different SU(2) multiplets and have different U(1) charge. Why are they all unpartnered and charged?
There is a simple reason for this--- any field which can get partnered will have an arbitrary mass term in the Lagrangian, and this term, without fine tuning, will end up generically being of order the Planck mass. So the only Fermions we see at low energies are those which are forbidden to have a mass, and therefore are chiral fermions without a partner to make a mass term with.
Further, all the fermions we see at low energies need to have a gauge charge, because without a charge of some sort, the Fermion can get a Majorana mass even without a partner, just by mixing with it's antiparticle. This is only forbidden if the particle is gauge charged in some way, so that the antiparticle has the opposite charge and the Majorana mixing is forbidden.
So all the fermions are chiral fermions with no partner to make a mass, so none of the low energy theories are vector-like.
The simplest right way to formulate gauge theories in a parity violating universe is in terms of 2-component spinors, each with an independent coupling to a collection of gauge fields. This procedure can lead to an inconsistency, if there is an anomaly in one of the gauged symmetries, so there are global constraints on the type of chiral fermions and the representations they can be in. If none of the Fermions have a partner, then the theory is natural, meaning "naturally massless" and the Fermions can only get a mass from a Higgs mechanism. The naturalness arguments say that the Higgs mechnism must be the source of mass of all Fermions in nature.
But if the Higgs is a fundamental scalar, the Higgs itself can have a mass, and the naturalness argument fails for the Higgs itself. So there is the question of why the Higgs has an unnaturally light mass. This is the hierarchy problem.
What a great question OP! I have good news and bad news. The good news is that this exact same question is asked and answered in Quantum Field Theory, by Itzykson & Zuber, section 12-5-2. The bad news is that the answer is
If you introduce mass terms in non-abelian gauge theories by hand, the theory is non-renormalisable.
This means that one is forced to introduce the Higgs mechanism (or variations thereof, such as the Stückelberg mechanism), which for some people is rather inelegant (and plagued by problems of naturalness, etc). Oh well, that's the way the cookie crumbles.
Let me quote the first paragraph of the aforementioned section, so as to summarise the main point of the problem:
Is a gauge theory where mass terms are introduced by hand renormalizable?
In electrodynamics, the situation is favorable. After separation of the gauge
field into transverse and longitudinal components, the longitudinal part $k_\mu k_\nu/M^2$ which gives rise to the bad behavior in the propagator does not contribute to the $S$ matrix. This results from the noninteraction of longitudinal and transverse components and from the coupling of the field to a conserved current. In a nonabelian theory, none of these properties is satisfied. Longitudinal and transverse parts do interact, while the current to which the gauge field is coupled is not conserved. On the other hand, unexpected cancellations of divergences at the one-loop level make the theory look like renormalizable. This explains why it took some time to reach a consensus, namely, that the theory is not renormalizable. The way out of this unpleasant situation is to appeal to the mechanism of spontaneous symmetry breaking, to be explained in the next subsection.
Best Answer
I) At the perturbative/diagrammatic level of photon self-energy/vacuum-polarization $\Pi^{\mu\nu}$ , the photon masslessness is protected by the Ward identity, which in turn is a consequence of - you guessed it - gauge invariance. For the explanation in the setting of QED, see e.g. Ref. 1.
Fig. 1: A one-loop contribution to the photon self-energy/vacuum-polarization. More generally, the 'bubble' in the middle could be 'filled' with higher-loop contributions.
A brief oversimplified explanation is as follows: Mass is associated with a Feynman diagram in Fig. 1 and its higher-loop counterparts. The Feynman diagram is built from Lorentz-covariant tensor objects. The Ward identity states, loosely speaking, that the photon 4-vector $k^{\mu}$ is perpendicular to the Lorentz tensor structure of the middle bubble part of the diagram. In the end only the bare propagator/tree diagram without any loops/bubbles survives, thereby rendering the photon massless.
II) It should perhaps also be mentioned that in the Higgs mechanism, the fact that the Higgs field $\phi$ transforms in the fundamental representation of the electroweak gauge group $SU(2)\times U(1)~$ leaves one of the four gauge bosons without a massterm in the Lagrangian - the photon, cf. e.g. Ref. 2.
References:
M.E. Peskin & D.V. Schroeder, An Intro to QFT, Section 7.5.
M.E. Peskin & D.V. Schroeder, An Intro to QFT, Section 20.2.