Why does a Pewter mug keep a beverage hot better than a foam cup? I know my subject sounds improbable and that's why I posted it.  I helped my daughter with a science project for school.  We tested five different materials to determine which would keep hot water hot, glass, ceramic, plastic, pewter and foam.  To may amazement, the pewter did the best, though only slighter better than foam.  I was so surprised I re-did the tests for the foam and pewter and got the same results.  We placed boiling water in the cup and measured the temperature over a five minute period.  The temp in the foam cup dropped about 24 degrees and in the pewter it dropped about 23 over the five minute period.  All the others were a degree or two over that.  I can't explain this and I am a chemical engineer.  Can't find much about it on the Internet either.
 A: The cup that insulates the best will minimize convention and present the hottest liquid at the surface, where the delta T between liquid and air will determine cooling.
The cup that conducts heat the best will carry heat from the top to the bottom and cause convection currents. The heat will spread more evenly throughout the liquid and the surface liquid will have a slightly lower temperature than with the best insulating cup. The delta T between liquid and air will be lower, even though heat content of the two cups of liquid were the same.
I would expect the insulated cup to cool faster if the mouth is the same size and shape and the liquid level is the same. A small change in air currents will have a big effect. Can you test with a lid? A floating disk of foam to fit each cup would do, and make measurements for 30 minutes.
That's my theory and I'm sticking to it!
A: The metal mug will equilibrate with the water much faster than the foam mug will— but after that the heat has no place to go$^\dagger$ except to be transferred away through radiation or be lost as steam via natural convection (which does not depend on the material of the mug). It turns out that radiation plays a very small role as shown below:
Say each mug is a cylinder with diameter $10\ cm$ and height $10\ cm$. Inputting the temperature of boiling water ($373\ K$) and the surface area of the mug ($150 \pi\ cm^2$) into the Stefan-Boltzmann Law gives an approximation of the heat radiated away by the cup of 15 Watts. 5 minutes at 15 Watts = 4,500 J of energy radiated away, which we will call $Q$. Assuming you filled the mugs to the top, there is $250 \pi\ cm^3$ of water or a mass $m$ of $250 \pi\ g$. Since water's heat capacity $C$ is $4.18\ J/g\cdot K$ you can calculate the temperature drop $\Delta T$ due to radiation using the definition of heat capacity:  
$$\Delta T= \frac{Q}{mC} =\ \frac{4,500\ J}{\left(250 \pi\ g\right)\left(4.18\ J/g\cdot K\right)}=1.3\ K$$
That is much less than the 23 degrees temperature drop you report, so the majority of the heat loss comes from evaporative cooling as suggested by Martin Beckett. That means the cooling rate is relatively unaffected by the material of the mug and instead depends mostly on the area of the water surface exposed to air. A great teaching opportunity about the different mechanisms of heat transfer!

$^\dagger$This is assuming no additional conduction to the table. If the table was made of metal you might see the effect you were expecting, since the heat would be able to conduct quickly through the pewter mug into the table.
A: The heat loss from an open cup was probably dominated by evaporation so the material didn't have very much effect and was within the limits of your experiment.
