If we define "temperature" as "a useful number proportional to the mean energy stored in some degree of freedom" (which is not the most fundamental definition of temperature, but it's close in this circumstance), then it's possible for the "translational temperature" and the "spin temperature" of a gas to get out of thermal equilibrium with each other. When this happens, a "thermometer" which couples to the translational degree of freedom will measure the translational temperature, while a thermometer which couples to the spin degree of freedom will measure the spin temperature. In general it takes more cleverness and subterfuge to build a "spin thermometer," which is where the sense of your quote comes from.
A naturally-occurring example of such disequilibrium occurs in diatomic hydrogen, $\rm H_2$. The two protons in a hydrogen molecules must obey Fermi-Dirac statistics, so their wavefunction must be antisymmetric under exchange. The rotational wavefunctions with quantum number $L$ have symmetry $(-1)^L$ under exchange — that is, the $s$-wave and $d$-wave are symmetric, the $p$-wave and $f$-wave are antisymmetric, and so on. To have the total wavefunction antisymmetric, the states with even $L$ must have the two protons in a spin singlet, while the states with odd $L$ must have the two protons in a spin triplet. The energies of the rotational states are
$$ E_L = \frac{15\rm\,meV}{2}L(L+1)
$$
Here the energy of the $L=1$ state, a few milli-eV, is consistent with the moment of inertia for a rotor whose mass and size are set by the hydrogen molecule's mass and bond length. There isn't any important correction to the molecular energy due to the proton-proton spin interaction because the proton's magnetic moment is very small. If the hydrogen gas is in thermal equilibrium, the probability of finding a molecule in a state with angular momentum $L$ is proportional to
$$
P(L) \propto (2S+1)(2L+1) e^{-E_L/kT}
$$
This is just the ordinary Boltzmann factor $e^{-E_\text{state}/kT}$ and the degeneracy factors for the spin and orbital angular momentum quantum numbers: there are $2S+1=3$ ways to orient a spin triplet, but only $2S+1=1$ way to orient a spin singlet. At room temperature ($kT=25\rm\,meV$), there are nontrivial populations in states with many $L$, so hydrogen's heat capacity is $\frac52R$. Near its triple point ($T \approx 30\,\mathrm K = 2.5\,\mathrm{meV}/k$) the rotational degrees of freedom are "frozen out" as the Boltzmann factor $e^{-E_L/kT}$ becomes smaller and smaller for all the states with $L>0$. The heat capacity of cold hydrogen gas approaches $\frac32R$ as the rotational states become less accessible.
At least, that's what you'd think until you starting making liquid hydrogen.
It turns out that there isn't any internal mechanism for an isolated molecule of $\rm H_2$ to convert from a spin triplet to a spin singlet (or vice-versa), but collisions between molecules allow spin exchange pretty effectively. So hydrogen which has been at room temperature for a long time will have a thermal-equilibrium distribution of $L$. The requirement that protons obey Fermi-Dirac statistics, and the spin-degeneracy factor $(2S+1)$ attached to the Boltzmann factor, conspire to mean that three-quarters of the population will be spin-triplet molecules with $L$ odd. When you cool hydrogen vapor down below its triple point, you actually don't bring all of the molecules into the $L=0$ ground state. Most of your hydrogen has odd $L$ and gets stuck in the $L=1$ state, waiting to collide with another $L=1$ molecule to be able to release its rotational energy. The even-$L$ and odd-$L$ molecules actually behave like two miscible but distinct gases, called "parahydrogen" ($L$ even) and "orthohydrogen" ($L$ odd). They have slightly different heat capacities, because the first excitation in parahydrogen $L=0\to2$ has different energy and degeneracy than the first excitation in orthohydrogen $L=1\to3$. They also have slightly different densities and other thermal properties.
The rate of ortho-to-para conversion in cold hydrogen (where the $L=1$ state should be "frozen out") is proportional to the square of the orthohydrogen density: the orthos have to find each other. When you liquify cold hydrogen vapor its total density increases by a thousandfold, and the rate of orthohydrogen downconversion increases by a factor of a million. Each $L=1\to0$ downconversion releases 15 meV, which happens to be approximately the same the latent heat of vaporization. So what happens when you liquify hydrogen gas from room temperature is
- Your room-temperature "normal hydrogen" (75% ortho, 25% para) falls below its boiling temperature and condenses into a liquid.
- The orthohydrogens in the dense liquid undergo collisions with each other more rapidly than they did in the sparse gas. They find each other, have spin-exchange collisions, and convert to parahydrogen.
- The conversion to parahydrogen releases so much heat that all of your hydrogen boils again!
- If you have kept your parahydrogen vapor in your cryostat, rather than venting it and replacing it with more normal hydrogen from your bottle outside, you can re-liquify the parahydrogen and it will stay liquid this time.
Any "thermometer" that you attach to this system is going to tell you that hydrogen's triple point is about 33 K and its boiling point at ordinary pressure is about 20 K; ordinary thermometers are not sensitive to the extra heat stored in orthohydrogen's orbital angular momentum degree of freedom. You cannot tell the difference between cold $L=0$ hydrogen gas and cold $L=1$ hydrogen gas using only a thermometer; this statement directly addresses your question.
The problem with the statement you quote, alluded to in the very first sentence of this long answer, is that thermometers don't measure internal energy. A thermometer measures temperature, which is is a system's "willingness to give up its internal energy," to paraphrase Schroeder's excellent intro-thermal textbook. In many systems the temperature and the internal energy are proportional to each other. But it's quite possible for a system with non-interacting degrees of freedom to function at a different temperature for each type of interaction. The most common examples are "spin temperatures," but there are others as well.
