Specification sheet claims are usually "best case" with optimum vin vout and power. The conditions you require do not match the converter's optimum operating conditions.
You are seeking to achieve 6V across 3 Ohms.
Power = V^2/R = 36/3 = 12 Watts.
At eg 12V in you you would need eg ~= 1.3 A average input at 75% efficiency to get 12 W out - and you could easily achieve 12 Watts output.
At 6V in you'd need 2.6A average and still doable.
At 4V in you need 3.9A average in and with duty cycle considerations you are near or above the limit for 12W out depending on overall achieved efficiency.
A single LiIon 18650 cell has a max of 4.2V, average operating voltage of 3.6V and useful voltage range of about 3V to 4V.
Let's see what we can expect.
The LM2587 has a 5A peak internal switch.
Duty cycle Toff:Ton ~~~= Vin: Vout/efficiency
Say 3.5: 6/75% = 3.5 :8 ~= 30% off, 70% on.
Max switch current = 5A.
Available Iin avg ~= Imax /2 x ton/tcycle
= 5/2 x 70% = 1.75A
Power in max = 3.3V x 1.75A = 5.8W
Power out max = Power in x efficiency
Say 5.8 x 75% = 4.3 W
Available V into 3 Ohms .
Power = V^2/R
Or V = sqrt(Power x R) =
= (4.3 x 3)^0.5 = 3.6V
You are getting somewhat better - converter is presumably operating at better efficiency than the 75% I used.
BUT, while the converter is capable of providing more power under the best Vin, Vout and load combinations, it falls somewhat short in this case.