If the meter doesn't measure true RMS correctly, including any DC offset, then it is not a true RMS meter. There is no point calling something what it isn't.
The terms one finds used to describe the capabilities of available meters are "True RMS" and "True RMS AC+DC". Perhaps "True RMS" should be "True RMS AC only", but in the year 2012 it's clear by inference that when they say "True RMS", they really mean "True RMS AC only".
There is a point in describing a meter as "True RMS" even if it's not able to include the DC component of a waveform in its measurement when on the AC range of the meter. This because it's possible to get the "True RMS AC+DC" value of a waveform from such a meter by measuring the DC component with a separate measurement and doing the V
AC+DC = sqrt(V
AC2 + V
DC2) computation.
If the meter is a "True RMS AC only" (called "True RMS" by manufacturers), this two part measurement is capable of giving the "True RMS AC+DC" value of the waveform. If the meter isn't even "True RMS" (meaning that it's an average responding meter), that measurement won't be possible at all. That's why there's a point in describing a meter as "True RMS"; we need only understand that it's a marketing gimmick.
Now, it would be my preference that the very designation RMS should mean a value for a waveform that includes the total waveform in the measurement/computation, AC and DC components both. But, in fact, because of marketing practices, some meters are only capable of measuring the RMS value of the AC part of a waveform when on the AC range. Such a meter can still measure the "True RMS AC+DC" value of a waveform with two measurements, which an average responding meter can't do. The buyer should understand what the manufacturers mean when they describe a meter as "True RMS" or TRMS, without the "AC+DC" addendum.
A good example of where it is absolutely necessary to be able to measure an RMS value which includes the DC component of the waveform is determining copper losses in a full wave capacitor input filter power supply, where the transformer secondary is center tapped (a two diode configuration). The current in each half of the center tapped winding has a substantial DC component because there is a single diode in series with the half winding; the current in that half winding is therefore unidirectional. With a "True RMS AC only" (called "True RMS" by manufacturers) you could make the measurement; with an average responding meter, you couldn't.
This means that it's not a good idea to use a 0->5v square logical signal to test it.
Such a waveform will work just fine; simply put a 1 uF film capacitor in series with it to eliminate the DC component of the waveform.
The answer to your original question is that probably the simplest way to determine if a meter is truly responding to the RMS value of the AC part of a waveform is to use a square wave without a DC component. If you have a bipolar square wave, with a positive excursion of 1 volt and a negative excursion of -1 volt, the RMS value of that waveform is 1 volt RMS, regardless of the duty factor. An average responding meter will calculate the average value of the absolute value of that waveform (the full wave rectified version) and multiply by the number 1.1107, which would give the correct RMS value of an undistorted sine wave. That same average responding meter would give a value of 1.111 volts for the square wave I described.
If you use a logic square wave with a maximum positive value of V volts (5 volts maybe), minimum value of zero volts, and remove the DC component with a capacitor, you will have a DC free square wave with a peak value of V/2 volts (2.5 volts maybe). A "True RMS AC only" meter will read that waveform as 2.5 volts; an average responding meter will read 2.5*1.111 = 2.777 volts.
You could even use the CAL output from a scope. Put a big film cap in series with it and measure with your meter. If the P-P voltage of the calibrator output (as measured by the scope) is V volts, then a "True RMS AC only" meter will read V/2 volts, but an average responding meter will read V/2*1.111 volts.
NOTE WELL. This assumes that the duty factor of the calibrator out is a 50% duty factor square wave. If it's not, use the following to compensate. (When you remove the DC component of a square wave, the positive and negative excursions change with changes in duty factor.)
Assume we have a square wave with a positive excursion of 1 volt and a negative excursion of -1 volt, with adjustable duty factor. If the DC component is removed with a capacitor, and the RMS value of the DC free waveform and the average of its absolute value (what a non RMS meter would measure) are calculated, we would get table shown in the first image. The first column is the duty factor in percent. The second column is the RMS value and the third column is the average of the absolute value. If you have a square wave with peak values other than 1 volt, use numbers in the table as compensating factors.
The second image shows the same information in graphical format with the duty factor along the horizontal axis. The red curve is for the RMS value and the blue curve is for the averaged absolute value.