For this you do not need fan speed control provided by a mobo fan header. You really want simple continuous operation of that fan from a fixed power source, such as is available from the PSU through existing wiring into the drive case; however, you want the speed reduced for noise reduction, so 12 VDC is too high a voltage. Regarding colour codes, from what you describe, the wires from the new fan motor should be: Black = Ground, Red = + DCV supply, Yellow = Speed signal, Blue = PWM Signal. For your purpose with no speed control or read-out, ignore and tape up the Yellow and Blue lines.
You have two ways to get reduced voltage to this fan. One simple way is to take slightly non-standard supply connections from a standard female 4-pin Molex output connector from the PSU. On them the two central lines are Black and grounds, Yellow on one end is +12 VDC, Red on the other end is +5 VDC. Whereas a common use is to connect between Yellow and Black for 12 VDC, using Yellow and Red gets you a difference of 7 VDC, sufficient to start and run a 12 VDC fan at reduced speed. To do this, connect Molex Yellow to Fan Red, and Molex Red to Fan Black.
The other way is to select the proper resistor and use the normal 12 VDC (Yellow and Black) connection from the Molex output. The voltage you supply to the fan should never be less than 5 VDC, and to be sure of start-up it really should be at least 7 VDC. Your "randomly-selected" resistor probably was just not the right value. So here's how to do the calcs, and there are two. One is the correct resistance value, the other is the proper wattage rating so the resistor can handle the heat it generates. Your fan draws at max 0.25 A when already running if fed a full 12 VDC, but more very briefly at start-up. So treat the motor as having a resistance of 12/0.25 = 48 Ohms. The added series resistor will have value R, yet to be found. We can set up two equations for your requirements.
Total resistance is R + 48 Ohms, total voltage across that is 12 V, current is unknown with value I.
Now the voltage is split with 7 volts drop across the motor at a current of I with a resistance of 48 Ohms, and a drop of 5 V across the resistor of R Ohms and a current of I.
So, I= 7/48 = 0.1458 amps from the motor,
and thus R=5/0.1458= 34.29 Ohms.
Just to check, if we put 48 Ohms (motor) and 34 Ohms (resistor) in series with 12 VDC across them the current will be 0.1469 amps, and the two voltage crops will be 7.02 V across the motor and 4.99 V across the resistor. The power dissipated in the resistor will be 4.99 x 0.1469 = 0.733 Watts.
So in rough terms you could use a resistor of 35 Ohms, with a 1 W rating.
Re-doing this for a 9 VDC supply to the motor (rather than 7 VDC), we get I= 9/48 = 0.1897 Amps, R = 3/.1875 = 16 Ohms, and resistor power rating = 3 * .1875 = 0.56W.
So use those two calcs as a "bracket". The Resistor value needs to be in the range 16 to 35 Ohms, with a power rating of 1 W. Use the lower end if you want the fan speed to be higher, and MAYBE use a 2 W resistor to be sure it does not get overheated. The real truth is that the effective "resistance" of the motor will not be as calculated above, because its real impedance is less that those values when it is running slower. But this range of Resistor value still will work.