If a module has only two terminals, but has some parallel LEDs internally, you can probably assume there's something inside the module to balance the current between each parallel circuit. This is made somewhat easier when all the LEDs are in one module because they can share a heatsink, and thus the temperature of each is likely close, and temperature variations are a big factor in why parallel LEDs will not share current equally. Each LED was also likely manufactured at the same time, so have very similar characteristics, which you can't guarantee when taking discrete LEDs out of a jar. The module may also include some small resistance in each parallel circuit to help balance. Point being: assume the manufacturer has taken care of this, and all you need to worry about is supplying the correct current to the two terminals provided by the module.
Your particular module is really like three modules, one for each color. It looks like there is one common terminal on the bottom, and three separate terminals on the top, one for each color. I can't find a datasheet that specifies if the cathodes or the anodes are common, so you may have to figure it out for yourself. It looks like maybe you can cut the common terminal apart if you want, but again, I see no datasheet, so you might have to experiment for yourself.
There is no special kind of LED driver that can drive parallel LED circuits. If an LED driver's job is to pump electrons, there's no way for it to tell some electrons to go down one circuit while telling others to go another way. The electrons decide which way to go by going whichever way minimizes their potential.
So, what you want to do with this module is power each of the R, G, and B sub-modules with a suitable driver (or if you can find it, one box that actually has three drivers in it). What you don't want to do is try to put the R, G, and B sub-modules in parallel and drive them all together. Since each color has a different forward voltage, this won't even remotely work: the color with the lowest forward voltage (red) will take very nearly all the current and all the power, and possibly be destroyed. At best you just won't get the other colors to light.
LEDs with different forward current requirements can not be connected in series. Doing so will cause some LEDs to be damaged.
LEDs with different forward voltages can not be connected in parallel. Doing so will cause some LEDs to be dim or completely dark.
LEDs that have the same current requirement can be connected in a series string, with an appropriate resistor to limit current or using a constant-current source.
LED strings can be connected in parallel as long as they are all designed to operate at the same voltage. That is, the current limiting resistor in each string must have been chosen for operation with the same voltage. So, connecting several strings in parallel and driving them with a constant-current driver is not likely to work well...they are not guaranteed to share current equally.
Best Answer
I'll address a single LED chain first, and get on to multiple chains later.
With a single chain of LEDs, powered by a current source, you do not even need resistor R2 in your circuit. The function of that resistor is to develop ("consume") whatever voltage is not required by the LEDs. For instance, if you employed a fixed voltage 18V supply, and your chain requires 9.6V, then the resistor is required to "absorb" 8.4V of the total 18V, so the remaining 9.6V is delivered to the LEDs.
By contrast, a 20mA current source (such as your LM317) will adjust its voltage to whatever value is required to pass exactly 20mA through whatever load is connected. If that load happens to develop 9.6V (like your 3 × LED chain), then that will be the exact voltage output by the source:
simulate this circuit – Schematic created using CircuitLab
On the right, where I use a fixed 18V supply, R1 is sized to develop \$18V - 9.6V = 8.4V\$ when passing 20mA, according to Ohm's law:
$$ R = \frac{V}{I} = \frac{18V-9.6V}{20mA} = 420\Omega $$
On the left above, where the power supply is a current source, no resistor is required, since the current source automatically produces 9.6V; that's what the diodes' total voltage would be when passing 20mA.
That means your own circuit with the single LED chain is incorrect; R2 would not be required. Do not confuse the purposes of R1 and R2. R1 in your circuit is used to tell the LM317 how much current to pass, and otherwise plays no role in limiting current by "consuming" unwanted voltage. Rather, R1 is calculated according to the following, assuming \$I\$ is the desired current output from the LM317:
$$ R_1 = \frac{1.25V}{I} = \frac{1.25V}{20mA} = 62.5\Omega $$
simulate this circuit
The LM317 is responsible for "dropping" whatever voltage is left over, after the total LED voltage and the voltage across R2 (1.25V) is removed from the supply. You won't need another resistor.
All that assumes you have one chain of LEDs, but your application requires 9 chains. This requires some extra understanding. In the following two circuits I power two chains of LEDs, from a current source providing 40mA:
simulate this circuit
On the left, all diodes are identical, well matched in all their parameters. As you can see, current is shared equally between both paths; 20mA each. That's the ideal scenario.
On the right, though, I've changed one of the diode's characteristics by a very small amount, and the result is that current is no longer shared equally. This is the problem with diodes, unless they are are all identical in every respect, they will have slightly different "forward voltages", and this will cause a large difference in how current is distributed through them, and consequently one chain will glow more brightly than the other.
The easy solution is to give some "elasticity" to the system, using small-value resistors, known as "ballast". Their purpose is to take up some of the voltage difference, giving each chain some flexibility in choosing its own total voltage. Here's the same (right hand) circuit reproduced with some ballast in each chain, which causes currents in each path to be closer in value:
simulate this circuit
All these facts together mean that you should use ballast resistors (one for each chain), even if you use a current source. Using an LM317 as a current source to power multiple chains of LEDs, while mitigating LED mismatches, will look like this:
simulate this circuit
If you have 9 chains, and each chain requires 20mA, then total current will be \$I=9 \times 20mA = 180mA\$. R1 will be calculated as follows:
$$ R_1 = \frac{1.25V}{180mA} = 6.9\Omega $$
There are caveats. R1 "consumes" 1.25V. Each ballast resistor "consumes" \$20mA \times 47\Omega = 1V\$. The LM317 regulator has a drop-out voltage of 2.5V. The LED chains each have 9.6V across them. The minimum supply voltage at the input to the LM317 will be the sum of all those:
$$ V_{SUPPLY} = 9.6V + 1V + 1.25V + 2.5V = 14.4V $$
This is bad news if you are planning on using a 12V supply. If this is a problem, but you are sure your voltage source will be fairly stable, then it might be simpler to use a single resistor for each chain, and forego the current source altogether, as in the right-hand circuit at the very top of this answer.