First, I'm suspicious of the specs. 12V 40W should be about 3.33A, but 4s7p suggests each LED is seeing 3.33A/7=476mA. I don't know that much about COB arrays, but I was under the impression they mostly used 1W dice, which would be limited to about 350mA. So I think maybe you really have a 28W array, that should only be driven at 2.5A. But I'd welcome correction from people who know more about these.
Second, yeah, that's being overdriven, and you will need some circuit to drive it right. Of course if you want to overdrive it, that's OK too -- just make sure you've got good cooling, and monitor the current as it heats up, to make sure it doesn't go into thermal runaway. See, as the LEDs get hotter, they'll take more current at the same voltage, thus burning more power, thus getting even hotter, and the cycle continues till the LEDs are ruined -- unless you have a good enough heatsink that things reach an equilibrium first. Resistors decrease this tendency, and proper constant-current drivers eliminate it altogether. (The LED can still overheat, but it won't receive ever-increasing power as it does so.)
Could be as simple as a resistor, but you don't have much headroom for that resistor, which means two things: you'll have poor regulation (large difference in brightness based on small fluctuations in battery voltage), and poor resistance to thermal runaway. The resistor moderates an LED's tendency to thermal runaway by reducing the voltage to the LED array as current increases, but this reduction depends on the resistance. The more difference between the LED voltage and the battery voltage, the bigger the resistor you'd use, and the more resistant to thermal runaway your system will be. (But also the worse efficiency will be, because the resistor just burns power off as heat.) If you go this route, you need to go from, let's say, 13.85V input to 12V output, with a current of 3.33A. So your resistor value is (13.85-12)/3.33 = 0.55Ω, and the resistor will dissipate about 6W, so it should be rated for at least 10 watts. Or maybe a 0.75Ω resistor, if you think it's really a 28W array.
A better choice for a low-headroom situation is a linear driver -- this burns the excess voltage off as heat, just like a resistor, but doesn't let the LEDs pull more current, so thermal runaway can't happen. Also the current remains constant regardless of voltage fluctuations. In your particular case, I'd use an AMC7135-based driver -- these are simple and cheap. The AMC7135 chips each drive 350mA (or 380mA, there's two flavors), so you'll need a driver with 8 chips (8x0.38=3.04A) or so. The only gotcha is that the maximum input voltage is 7V, so you'll need to provide a 5V (or so) input for the AMC7135s' control signal (less than 1 mA). You can use a voltage regulator, a zener diode & resistor, or a pair of resistors acting as a voltage divider. These can't run with more than 2.5V difference without extreme heatsinking measures, but in your application, they should be OK; just stick the whole driver to a good heatsink with whatever sort of thermal adhesive. You could use one like this from RMM's shop, with 8 chips on one driver (choose the single-mode option), or two of these 4-chip drivers from dx in parallel. (Also available a bunch of other places.)
Finally, you could use a switching power supply -- in this case, a buck converter, not a boost converter. But these are typically 80-95% efficiency, so when your supply voltage is less than 20% over your output voltage, they may not be any more efficient than a linear driver.