Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.
What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?
There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are vacant), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is accessible).
It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.
As Nicolau hints in his answer, orbitals are not regions of space around an atomic nucleus. They are mathematical constructs formulated as wavefunctions describing the properties of electrons, including their energy, angular momentum, and probabilistic distribution in space. The pictures of "orbitals" that you are used to seeing are the probability density functions (the absolute value of the wavefunction squared).
Ultimately, the only way we can know what a probability density function looks like or what the energy of orbital would be is to pretend there is an electron in the orbital and then do the math. So, orbitals without electrons in them might as well not exist. As Nicolau and ManishEarth point out for oxygen, the orbitals in the 3d subshell are basically inaccessible to the electrons in oxygen. Thus those orbitals do not exist.
Oxygen has a d-orbital as well. Just that there are $3s$ and $3p$ orbitals in between, so extra electrons will get filled there first. The $3d$ orbital has too high energy for it to be of any use.
Also, hybridization almost always (the exception is with coordination compounds) happens for orbitals with the same principal quantum number (shell number). So the $3s$ and $3p$ orbitals are not of any use for accomodating extra electrons/bonds.
Orbitals are uncorrelated spatial quantum states of electrons. This representation makes it intelectually more confortable and allows one to approximate the Schrödinger equation. Nevertheless, for polyelectronic atoms one have to correlate the electrons in order to get to the right energy. On way to do so is to construct a wave function that implies all the orbitals : the one that are occupied (in the fondamental state) and the virtual ones (unoccupied in the fondamental state). For O and S, 3d orbitals are virtual orbitals. By implying those states in the wave function we are able to correlate the electrons and get the right energy for the hole system. Quantum mechanics says that the electron is in all the states at the sametime but with a different probability and the probability to find an electron in 3d orbital for O is much less than for an S electron.
Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.
The most weakly bound electrons in oxygen are in the 2p orbital using a one-electron approximation. There is no 2d orbital, so the next highest orbital is the 3s orbital. You could have an excited state where the 3s orbital gets populated. However, there are no compounds of oxygen where electrons are in the 3s orbital or "make use" of the 3s orbital (in the sense of hybridization or linear combination of atomic orbitals).
The most weakly bound electrons in sulfur are in the 3p orbital using a one-electron approximation. There is a 3d orbital with higher energy, and you could have an excited state where that gets populated. "Making use" of the 3d orbitals has been invoked in compounds of sulfur (like $\ce{SF6}$) but does not really happen either, see e.g. Hypervalency of elements.
What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?
Orbitals are solutions (eigenfunctions) of the Schrödinger equation. There is no solution that corresponds to 2d (that would be n=2 and l=2), but there is one that corresponds to 3d (that is n=3 and l=2)
The principle quantum number n (e.g. 2 for the 2p orbital or 3 for the 3p orbital) gives you the total number of the spherical and planar and conical nodes plus one. The d orbitals have two planar or two conical nodes, so the principle quantum number has to be 3 at least.
