Why do the Wh and mAh ratings of a USB battery pack appear to be inconsistent?
Firstly the battery is not \$5\mathrm{V}\$, it is nominally \$3.7\mathrm{V}\$ as pointed out already.

However the confusion lies in misinterpreting the \$\mathrm{mAh}\$ rating of the battery, and it has nothing to do with converter losses, and everything to do with

$$P=I\times V$$

If the battery is rated at \$6\mathrm{Ah}\$, it means you can draw \$1\mathrm{A}\$ for \$6\mathrm{\space hours}\$. So lets say you are drawing \$1\mathrm{A}\$ from the battery. This means it is delivering:

$$P=I\times V = 1 \times 3.7 = 3.7\mathrm{W}$$

Now lets say you feed this through an ideal boost converter. In this case \$P_{in}=P_{out}\$. So assuming we boost it up to \$5\mathrm{V}\$, this means that the current that must be drawn from the output to discharge the battery at this rate is:

$$I=P/V=3.7/5=0.74\mathrm{A}$$

So what this means is at the higher voltage, you can draw a current \$0.74\mathrm{A}\$ for \$6\mathrm{\space h}\$ for that battery capacity - so the output capacity is \$4.44\mathrm{Ah}\$.

Again, this is not converter losses - in fact with losses the number would be lower. Instead it has to do with the fact that you are sacrificing current capacity for a voltage gain which is how a boost converter works - if this wasn't the case then you would have invented free energy. 

Essentially at a higher voltage, the energy delivered by each unit of charge \$\left(\mathrm{V}=\mathrm{J}/\mathrm{C}\right)\$ is higher, hence at a higher voltage but lower current you are still delivering the same energy.