If you want to remove noise from a power supply rail, you would be bettern off using a series inductor followed by a capacitor to ground.
If you use a series resistor in a power supply rail, as soon as you start to draw current from it, the resistor will start dissipating large amounts of the voltage. take your 220R resistor for example - at 10mA, it would drop nearly half the +5v supply (2.2v). An inductor by comparison has negligible resistance (generally) and will do a better job at filtering.
I have used a cheap 220uH inductor and a 10uF capacitor to filter the power supply for an audio DAC with great success. This combination would give you a cutoff of:
1/(2pisqrt(L*C)) = 3393Hz approximately.
There are two things you may have also failed to account for. First, you cant just use any old capacitor to make an RC filter at 10MHz. Cheap capacitors tend to have a tollerance of +-20%, sometimes as high as 80%, which means you can't gaurantee the capacitance value and thus your cutoff point.
The second thing is 'cutoff' is a bit deceptive. What it means is the frequency at which the signal drops to half the magnitude (aka the 3dB point). It is impossible to get a brick wall filter as it is known (unless your signal can time travel), and so you have to allow for this in your filter. For simple filters like the one you have made has not been carefully designed to have as sharp a cutoff as possible.
For DC power supplies you aren't just trying to cutoff the high freqency component, as the noise is not just at the 15MHz your scope estimate, but across large parts of the specutrum including low frequencies. For a power supply, you want to remove as much noise as possible and only keep the DC component. This means you want your cutoff as close to 0 as you can, without attenuating the DC voltage.