So my question is if we have a pulse wave does it make sense anymore
to talk about its RMS value?
Yes, of course it does; the rms value of the pulse wave is the effective DC voltage across a resistor that gives the same average power.
Recall that the instantaneous power associated with a resistor is
$$p_R(t) = \dfrac{v^2_R(t)}{R} $$
The average power, over a period \$T\$, is then
$$p_{avg} = \dfrac{1}{T} \int_0^Tp_R(t)\,dt =\dfrac{1}{T} \int_0^T\dfrac{v^2_R(t)}{R}\,dt$$
Thus, the equivalent DC voltage that produces the same average power is
$$V_{eq} = \sqrt{p_{avg}\cdot R} = \sqrt{\dfrac{1}{T} \int_0^Tv^2_R(t)\,dt}$$
But, that last term is precisely the root of the mean of the square (rms) value of \$v_R(t)\$.
So, yes, it makes sense to talk about the rms value of a pulse waveform or any other voltage or current waveform for that matter.
Ohm's law
$$
1: V(t) = I(t)R
$$
Instantaneous power dissipation is product of voltage and current
$$
2: P(t) = V(t)I(t)\\
$$
Substitute 1 into 2 to get instantaneous power through a resistor in terms of voltage or current:
$$
3: P(t) = I^2(t)R = \frac{V^2(t)}{R}\\
$$
Average power is definitionally the integral of instantaneous power over a period, divided by that period. Substitute 3 into that to get average power in terms of voltage and current.
$$
4: P_{avg}=\frac{\int_0^T{P(t)dt}}{T}=\frac{R\int_0^T{I^2(t)dt}}{T}=\frac{\int_0^T{V^2(t)dt}}{RT}\\
$$
Definition of RMS current
$$
5: I_{RMS}=\sqrt{\frac{\int_0^T{I^2(t)dt}}{T}}\\
$$
Square both sides
$$
6: I_{RMS}^2 =\frac{\int_0^T{I^2(t)dt}}{T}\\
$$
Multiply by R to find equation 4 for average power
$$
7: I_{RMS}^2R =\frac{R\int_0^T{I^2(t)dt}}{T}=P_{avg}\\
$$
Definition of RMS voltage
$$
8: V_{RMS}=\sqrt{\frac{\int_0^T{V^2(t)dt}}{T}}\\
$$
Square both sides
$$
9: V_{RMS}^2=\frac{\int_0^T{V^2(t)dt}}{T}\\
$$
Divide by R to find equation 4 for average power
$$
10: \frac{V_{RMS}^2}{R}=\frac{\int_0^T{V^2(t)dt}}{RT}=P_{avg}\\
$$
Multiply expressions 7 and 10 for average power
$$
11: P_{avg}^2=V_{RMS}^2I_{RMS}^2\\
$$
Square root of both sides
$$
12: P_{avg} = V_{RMS}I_{RMS}\\
$$
Q.E.D.
Best Answer
The rms voltage is given by
\$\sqrt{\frac{1}{t_1-t_0}\int_{t_0}^{t_1}v^2(t) \mathrm dt}\$
The mean voltage is given by
\$\frac{1}{t_1-t_0}\int_{t_0}^{t_1}v(t) \mathrm dt\$
As you can see, these are not the same, except in special cases. A pure positive dc voltage is one such special case.
However, if the dc component of voltage is much bigger than any ac components, the rms and mean will be very close to each other. This could apply to the case of "noise with a huge offset".