Your source signal is in principle an AC + DC signal. Your pulse train is your AC signal, with an amplitude at a certain value, if your plot is accurate, about 220mV I'd say (half the energy up top, half below -> the virtual middle is a little below the exact middle of the signal).
It is then offset to a DC value, around which it toggles/oscillates. Again, if your plot is correct, approximately 700mV.
Your capacitor effectively blocks low frequency signals, or this DC offset, so you remain with only the AC portion, around 0.
Either your question is:
- How come it goes negative? Then the answer is: because AC signals swing around 0V, which means they also go negative.
- How come it takes time to go to the steady state? Then the answer is: Yes, because of the R-C time it takes to re-balance the output after you start "exciting" the input.
When you first start toggling the input, both plates of the capacitor will want to jump up, and the resistor is quite large, so it can't prevent that. Since at the start-up the output was at 0V, it jumps a full swing up on the first flank. Then after that slowly the output plate will start to balance around 0V again, with both positive and negative peaks.
You can also see this charging effect on the low and high steady periods of the signal: Small slopes that are the start of an RC curve. If your signal gets slower, you'll see more and more of that RC-charge curve.
Want it to go to the steady state faster? Smaller capacitor, or smaller resistor.
Want it to stay above 0v? You'll need to add some semiconductor action. For such small signals it's possible you need some transistor action even, but there are Schottky diodes with 50mV forward drop at very low currents. If you connect those like this:
simulate this circuit – Schematic created using CircuitLab
You may need a (small) resistor on the input side as well, but that (and its value) would depend on more specifics you haven't given.
This way, every time the output plate swings negative, the diode pushes it back up nearly to 0V.
If you want real 0V with such a small signal that'll take transistors and/or op-amps and a supply of at least 2V (or even more complication if that 2V is not available)