MattiasOfTheMetal:
I don't have a scope unfortunately. And I tried that, but it did the same thing. I think I just can't find the right cap/resistor combination.
On my breadboard I had it like this:
To pedaboard, Cap leg with resistor to ground, Cap leg to DC power, DC power, DC power to Cap leg, Cap leg w/ resistor to ground, to guitar;
and of course common ground between the AC signal and DC power. I tried with smaller caps, but it would just cut out sound, and larger caps in parallel let more sound through, but as soon as DC was applied there was no sound at all, minus an oscillating whine.
I revisited my calculation example below, but that should give you a value that will work. However, this might be a good place to recall that even if you don't have the exact value needed with a single capacitor, it's possible to get close to it with combinations of parallel and serial capacitors. Just remember that they combine differently than resistors, basically the rules for parallel and serial addition are reversed.
MattiasOfTheMetal:
I have been considering other cables. sigh I have so many mono 1/4 cables as backup, I'd hate to have to switch to another type.
As for a midi cable, that is a good idea; but the music I play, we tend to not have any use for midi..(YET! until I build a midi tracker for my guitar, but thats another story).
I have considered using a Stereo 1/4 cable. On the tip would be the guitar's signal, ring would have the ground, and the shield would have power.
I would swap the ground and power only if it wouldn't make it sound horrible. And all I would need is a short sensing circuit on my pedalboard, basically measuring the resistance between the DC and ground. The arduino would do a test on startup and make sure the resistance is >100 ohms, then enable the DC power. So it could sense a mono cable plugged in instead, and automatically disable the power.
But then I'd have to buy some Stereo cables, and they may be less common on stage.
Well you are the one who will be using it so ultimately you will have to balance all the factors and decide which wiring approach is the best, all things considered. What I do know is if the AC has it's own dedicated wire, high-pass filters or coupling capacitors shouldn't be necessary. 
MattiasOfTheMetal:
This is the math:
|
V _________
---/ d(0.0)b --->
^
Right over my head.. ha
Honestly it shouldn't be over your head. It's true deriving the the RC time constant formula involves calculus. However using it to get a capacitor value in this case only requires solving for a single unknown using algebra. This is only 7th or 8th grade math, and you've demonstrated a grasp of mathematical concepts at least as, if not more, difficult in this thread.
MattiasOfTheMetal:
How would I know which frequency to drown out for the AC? I may be wrong, but I though DC was a low frequency.
DC is basically zero frequency, but the confusion is really my fault. I'm an EE that doesn't work with audio much. Off hand, I do know that the range of human hearing is kilohertz in magnitude, but I didn't double check the exact range. So for my example in an effort to get something that would yield some fairly easy numbers, I ended up choosing a frequency for the high pass filter that would actually exclude the entire audiable range for most human beings. 
The math is still sound, but you probably want to try something around 20 Hz , 0.05 seconds, instead.
The corrected example:
? = R * C
? = 0.05 s (20 Hz, inverted)
R = 8000 ? (lowest possible load is the worst case)
Solve for C
C = 0.05 s / 8000 ? = 6.25 ?F
MattiasOfTheMetal:
How did you invert 20khz into 5*10^-5 ?
The snarky answer would be "I just hit the '1/x' button on a calculator." 
However a more meaningful answer is that the frequency and period are reciprocal for a time varying signal. Therefore inverting one will always give you the other. Yet, as mentioned above these exact values were entirely incorrect for what you want.
MattiasOfTheMetal:
And to what scale would I use the resistor to ground the cap?
If I've understood you correctly, when hooked-up to the guitar the load would be on a range between 8k and 25k ?. In this case you really shouldn't need an additional resistor if the load is directly between the capacitor and ground. Although using one will be necessary if testing the circuit unconnected to the actual load.
MattiasOfTheMetal:
Since it would be a cutoff for (above?) 20khz, would that alter all sound above it? (fm frequencies for communication between arduinos)
It's my understanding that we were discussing overlaying AC on the DC power, not on any of the wires going directly I/O (should never be done because the Arduino microprocessors can only withstand about -0.5 V on these pins). If that is correct, the communication between the Arduinos will be on another wire or using wireless radio. Therefore, barring electromagnetic interference (which can be mitigated if it occurs), this AC signal shouldn't affect the Arduino serial communication by either means.