What determines the spin of fields in gauge field theories? I understand that gauge bosons transform as the adjoint of their respective symmetry groups, but what determines the spin of the field? Can you have some gauge group where the adjoint is spin zero?
 A: As was already mentioned in comments, the question is ambiguous. In any case 'spin' is the same as (irreducible) representation of some group, i.e. specifying spin is equivalent to giving the lowest weight of some group or its Lie algebra. Now, there are at least three different types of spin one can think about after your question.
1) Spin refers to the tensor type of the field your work with. Here tensor means tensor of the Lorentz grop $SO(d-1,1)$ if we are in dimension $d$. Then $\phi(x)$ has spin zero, the Yang-Mills field, $A_\mu$, has spin one (and we ignore all other types of indices it can carry). Graviton, $g_{\mu\nu}$, has spin two.
2) Spin refers to the physical degrees of freedom, i.e. to the solution space of the field equations. The relevant group in this context is the Wigner little group, (see, e.g. the second chapter of Weinberg's Quantum field theory). For massive fields Wigner's little group is $SO(d-1)$ and it is $SO(d-2)$ for massless. To determine spin one has to solve linearized equations of motion. Then $A_\mu$ turns out to carry spin one, the number of spin one fields we have is given by the dimension of adjoint, i.e. just the number of $A_\mu(x)$'s in $A^a_\mu T_a$. In this context on needs background, Minkowski space in our case, that has enough symmetries at least assymptotically.
3) Spin refers to the representation of the gauge group. It does not necessarily have to be the adjoint. (This usage of spin is extremely misleading)
