Why is the entropy of deuterium greater than that of hydrogen gas
The entropy can be split into three parts arising from translational, rotational and vibrational energies and using statistical mechanics each of there can be calculated from measure properties. The gases $\ce{H2}$ and $\ce{D2}$ differ in each of these three energies due to their differing mass and nuclear spin. The experimental values are hydrogen $130.7$ and deuterium $144.9 ~\pu{J mol^{-1}K^{-1}}$

The translational entropy is given by the Sakur-Tetrode equation, 
$$S_{trans} = R\left(\frac{5}{2}+\ln \left(\frac{V}{N} \right)+ \frac{3}{2}\ln\left( \frac{2\pi mkT}{h^2} \right )\right)$$
where m is the mass, V volume, N Avogadro’s number, h the Planck constant and k the Boltzmann constant. Assuming that the gases behave ideally the difference in entropy is small as it depends only on the mass difference all other terms being the same, so $S_{trans}$ is greater for $\ce{D2}$ than $\ce{H2}$ by $(3R/2)\ln(2) \approx 8.64~ \pu{ J mol^{-1}K^{-1}}$.

The vibrational entropy can be ignored at low temperature, for example at room temperature $kT \approx 210~ \pu{cm^{-1}}$ whereas the vibrational quantum are $4395$ and $3118~ \pu{cm^{-1}}$ for $\ce{H2}$ and $\ce{D2}$ respectively.

The contribution from the rotational motion depends both on the mass, as the moment of inertia determines the rotational constant, and on the nuclear spin which via the symmetry properties of Fermions and Bosons affects the weighting of the rotational energy levels.

To calculate the entropy the partition function $Z$ is required. Once this is obtained the entropy is calculated as
$$ S = k\ln(Z)+\frac{U}{T}$$
where the internal energy $U$ is 
$$U= kT^2\left ( \frac{\partial \ln(Z)}{\partial T} \right)_V$$

The partition function for odd mass number, assuming a rigid rotor model, is (see this answer for details Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance? )
$$Z_{odd}= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$

where $g_n = 2s_n+1$ where $s_n$ is the nuclear spin, $1/2$ for hydrogen and $1$ for deuterium. The rotational quantum numbers are J and $B_e$ the rotational constant assumed to be the same for all J but which differs between hydrogen ($60.85 \pu{cm^{-1}}$) and deuterium ($30.44 \pu{cm^{-1}}$) by a factor of two. (These rotational constants are not typical, values for most molecules which are heavier and have longer bonds are far smaller, for example, $\ce{F2}, B_e =1 \pu{cm^{-1}}$)

The partition function for even mass number
$$Z_{even}= \frac{g_n(g_n+1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n-1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$
where the only difference is that due to the weighting (those terms in $g_n$ ) in from of the summations which run over even and odd values of J. Hydrogen has weightings $1$ and $3$, deuterium $6$  and $2$ respectively in the equations above.
Thus Z for hydrogen with $g_n=2$ is 
$$Z=1\cdot\Sigma (even~J) + 3\cdot\Sigma(odd~J)$$
and for deuterium, $g_n=3$
$$Z=6\cdot\Sigma (even~J) + 2\cdot\Sigma(odd~J)$$

At low temperatures the argument in the exponential is large meaning that the exponential is small. e.g. $exp(-large ~number) \rightarrow 0$  for all J terms except $J=0$ where the exponential is $1$. Thus in hydrogen the limiting partition function at low temperatures is $Z_{T\rightarrow 0}= 1 + 3(small~number) \approx 1$ and the limiting entropy is $S_{T\rightarrow 0 } R\ln(1) \rightarrow 0$; (and $U(0)=0$)

In deuterium the limit is $Z_{T\rightarrow 0 } = 6 + 2(small~ number) \rightarrow 6 $ and thus the limiting entropy is $S_{T\rightarrow 0 } R\ln(6) \rightarrow 14.9~ \pu{J mol^{-1}K^{-1}}$.
The change of rotational entropy with temperature is shown in the figure.