There is only one temperature that matters, and that is the junction temperature.
You could use a small thermocouple and try to put it under the base, perhaps by drilling a small blind hole into the heatsink under the LED base. (from the other side). Lots of inexpensive multimeters have type-K (Chromel-Alumel) thermocouple inputs that are probably good enough for your purposes provided you check that they're reading correct room temperature when cold.
Another method would be to pass a small constant current such as 1mA through the LED junction and vary the temperature in an oven from room temperature up to, say, 100°C.
Plot that graph, fit a curve to it, or whatever, so you can find the inverse (temperature from voltage at 1mA). Now if you run the lamp at a known ambient temperature, allow it to stabilize and then very quickly switch from the "on" current of 600mA or whatever to the 1mA current you can get a true reading of the actual junction temperature before it cools too much.
The above method is similar to the standard method of checking temperature rise in transformers- using coil resistance (though we can skip the oven step because we know pretty much how the copper wire will behave with temperature, we just need one resistance reading at one known temperature).
As an approximation, with your IR thermometer, you could paint the back of the heatsink black (if it isn't already) to make the emissivity closer to 1, measure the temperature and add a fudge factor such such as 3-3.5°C/watt of dissipation to account for the junction-package and package-heatsink thermal resistance (use the actual number for the junction-to-case \$R_{\theta JA}\$ if you have a datasheet for the LED module. There will be a bit of gradient through the thickness of the heatsink but if you can get right under it with just some copper or aluminum between, the number should be close enough.
Current LEDs are good enough that it's a bit pessimistic to assume that all the electrical power that goes in is converted to heat- a significant amount is emitted as light, so does not contribute to heating of the LED.
