Can someone provide a rule of thumb on how I can calculate  annual heat load based on known heat loss/hour at any given temperature?  I imagine there is a way to calculate it with that info plus heating degree days?

Design heat loss is the amount of btu’s per hour at the 99% design temperature. Lets say the design temperature where you live is 8° and the temp calculated for the house is 68° The delta is 60 ° So you can estimate the btu’s/ deg F by dividing heat loss by 60.
Now you have btu’s per hour per degree Multiply that by 24 and you get btu’s per day per degree.

Look up the heating degree days (HDD) for your location use base 65 as a first estimate. Multiply HDD * (btu’s per day per degree) and what is left is yearly BTU’s

This is a crude approximation, but it will get you in the ballpark . Generally the more efficient the house, the lower the base value for Heating degrees Days to be more accurate. This also ignores solar gain, and does not handle other internal gains well.
You also have to factor in that a manual J heat loss tends to overestimate actual heat loss by a bit.

Using base 65F works pretty well for code-min houses, but for higher performance houses you really need to drop it. Many PassiveHouses "balance" at about 45-50F using standard Manual-J calculations, but the solar gains are so high relative to the load that the energy use is much less than would be indicated.

BeOpt does a better job of estimating overall energy use, as well as the actual heat loads.

Thanks for that explanation.

So if it's 26k btu/hour @ -11 F (99% design temp). at +66F for interior temp.

26,000 / delta of 77F (+66F minus -11F) = 337.66 * 24 hours = 8103 btus per day per degree.. * 7000 HDD (taken from http://www.degreedays.net) = 56,727,272 btus/year .

Sounds roughly correct?

Yep! With a condensing propane boiler averaging 95% efficiency that would be about (92,000 BTU/gallon x 0.95=) 87,500 BTU/gallon, or about (56,727,272 / 87,500 =) 650 gallons/year.

With mini-splits averaging a COP of 2.8 (you might do slightly better than that) it would be 3412 BTU/ kwh x 2.8= 9500 BTU/kwh, or (56,727,272 / 9500= ) about 6000 kwh /year.

Even with solar gain skewing the energy use it won't be half that, and even with higher than anticipated air leakage it won't be 50% more than that.

In a way better than code house it's likely to have a significantly lower balance point than 65F, with lower energy use than those numbers imply. It's probably worth running the numbers at base 55F and base 60F to get a feel for the actual energy use to expect. It'll probably be higher than the simple calculation using base 55F, but could still be under the number calculated using base 65F.

calculate  annual heat load based on known heat loss/hour at any given temperature?

If you can achieve fairly steady state nighttime indoor and outdoor conditions and then accurately meter energy use for an hour or so, you can calculate a btu per delta degree (indoor - outdoor). Repeat a few times and it might be more accurate than an average Manual J estimate - I don't know, I haven't seen comparison data for such a method.

Existing Building Energy Usage Analysis Software

Integrated Heating System Performance Software

As others have indicated, a base of 55-65F is more appropriate for low-load, energy efficient buildings and manual J numbers will tend to overestimate actual heat load.