Though it does go against your intuition, you've actually mentioned the answer in your question. Stibane has a higher boiling point than ammonia/azane on account of van der Waals interactions (owing to the larger size of the antimony atom).
Our teacher had actually posed this question to us during my first year of high-school. All of us were incredulous when we were told that stibane had a higher boiling point than ammonia, and like you, we were fixated on H-bonding and chose to dismiss van der Waals' interactions as insignificant in this regard.
First, a look at the H-bonding argument:
At school we're taught that H-bonding only occurs in molecules where you find hydrogen atom(s) bound to either nitrogen, oxygen or fluorine ($\ce{N,O,F}$). This is because $\ce{N, O}$ and $\ce{F}$ are sufficiently electronegative (about 3.0, 3.5 and 4.0 on the Pauling scale respectively) to polarize the $\ce{N/O/F-H}$ bond. Do note, that nitrogen is only barely able to establish H-bonding, while it's much easier for oxygen and fluorine to polarize the $\ce{O/F-H}$ bond on account of 1) their higher electronegativites and 2) their smaller sizes.
As antimony is much larger than nitrogen, and since the hydrogen bonding due to the nitrogen in ammonia is pretty weak, the van der Waals interactions in stibane 'outdoes' the H-bonding in ammonia.
However the same argument does not follow for water and hydrogen fluoride, which have higher boiling points than tellane and hydrogen iodide respectively.
$$\textbf{Melting } (\vartheta_\mathrm{m})\textbf{ and boiling } (\vartheta_\mathrm{b}) \textbf{ points of}\\\textbf{Group 15, 16, and 17 hydrides in K}$$
\begin{array}{lrrlrrlrr}\hline
&\hspace{-2em}\text{Group 15}\hspace{-1em} &&& \hspace{-2em}\text{Group 16}\hspace{-1em} &&& \hspace{-2em}\text{Group 17}\hspace{-1em}&\\ \hline
\text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b}\\ \hline
\ce{NH3} & 195.5 & 239.6 & \ce{H2O} & 273.0 & 373.0 & \ce{HF} & 180.7 & 292.4\\
\ce{PH3} & 138.0 & 185.0 & \ce{H2S} & 190.3 & 211.0 & \ce{HCl} & 161.0 & 189.4\\
\ce{AsH3} & 159.0 & 218.0 & \ce{H2Se} & 209.0 & 231.0 & \ce{HBr} & 184.5 & 206.0\\
\ce{SbH3} & 184.0 & 256.0 & \ce{H2Te} & 222.0 & 271.0 & \ce{HI} & 222.2 & 237.0 \\ \hline\end{array}
Now here's where your logic comes into play; the H-bonding's pretty strong in case of water and hydrogen fluoride (after all, oxygen and fluorine are the two most electronegative elements we know) and this time 'round the H-bonding outdoes the van der Waals interactions.
So you see, you should've made a comparision between the higher and lower hydrides (for groups 15, 16 and 17, to really be able to spot the exception).
As for your spin-off question:
Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?
Well, the H-bonding here is strong enough to keep the ammonia molecules together at normal temperatures. But you must never forget that H-bonding is not a true chemical bond, but an intermolecular interaction. If you provide sufficient energy to a sample of ammonia (i.e. by heating it) the ammonia molecules wiggle and wiggle and wiggle, until at some point they wiggle so rapidly, that they overcome the attractive forces brought about by hydrogen bonding and become a gas. This is why you don't observe H-bonding in gaseous ammonia: because the particles are too energetic to be held down.
