The rise/fall times are not that critical, no. As shown in the data sheet for the device you reference:
Referring to Figure 33, the clock signal alternatively switches
the SHA between sample mode and hold mode. When the SHA
is switched into sample mode, the signal source must be capable
of charging the sample capacitors and settling within one-half
of a clock cycle.
So it is not how fast it switches from LOW to HIGH or HIGH to LOW, but how long it remains in that steady state, and how stable your signal is during that state.
Basically, when the clock is HIGH the SHA capacitor is "reading" the analogue signal. When the clock is LOW the SHA capacitor is isolated from the analogue signal, and its voltage is presented to the sampling pipelines for conversion.
Obviously you want to read the value from that capacitor before it decays too much, so an overly huge fall time could be detrimental, but in general the rise and fall times are dwarfed by the required minimum level times (6.2ns to 15ns depending on model). There is no stated maximum time, only a minimum sample rate of 1MSPS, so a maximum clock period of 1µs, or a rise + hold or fall + hold maximum time of 500ns (with the previously stated minimum hold time) by my calculations.
The SHA capacitor effectively looks a little like a low-pass filter, with a time constant equal to the SHA charge time. For one of the devices that is 6.2ns, but others are as much as 15ns.
So the value in the SHA capacitor is effectively like a weighted rolling average of the past 6.2ns of voltages, and the final value is taken at the moment the ADC switches from SAMPLE to HOLD. Therefore you should ensure that the falling edge if the sampling clock occurs during a period where your sampled signal is stable, as it's the end of the sampling period, not the start that gives you your final value.
From your given waveforms, the TRIG signal falls during a trough in the video waveform. The sampled value would therefore be influenced by that trough and the results would be low. The SAMP signal, as the inverse, falls during one of the stable 2.5V periods, and that has been stable for more than 6.2ns, so would give a much more reliable reading of the 2.5V level.
Best Answer
Unfortunately this isn't really a mechanical problem and has little to do with the relay itself.
A mercury-wetted relay is certainly the right choice, but their contact closure speeds are a couple orders of magnitude faster than the rise times you're seeing. They typically achieve contact closure times of 3-5 picoseconds, and the contact break time (at least, in the galvanic sense) is going to be on a similar scale.
The rise time as well as the longer fall time are hundreds of times slower than the actual physical contact closure/release times due to parasitics.
1. It's all about energy
Any transmission line (or any conductor for that matter) will have some inductance. Inductance is a measure of how much energy for a particular current something will store in a magnetic field. During the rise time, the rise in current will not occur at the speed of the relay contact closure, but rather the speed at which the parasitic inductance allows. The pulse at the relay will have much sharper rise times, but the pulse as seen at the other end of the transmission line will be slower, caused by the slowed current rise time. The current can't rise as fast as the voltage at the relay contacts would otherwise cause because the inductance represents reactance/imaginary component of impedance, which is just like resistance but represents stored (as opposed to dissipated) energy. It is still impedance, it is still correctly measured in Ω, and it causes a voltage drop. Of course, the contribution to the impedance from inductance (assuming a constant voltage across the inductive element) is temporary and falls to zero once the current (and the resulting magnetic field it creates) reaches a steady state.
2. You don't need much
And while a good transmission line will have very low inductance, it is important to understand just how little inductance is needed to cause something like a 700ps rise time: 10nH would do it at 250V.
On a circuit board that is 0.78mm thick with a solid ground plane on one side, any trace on the opposite side that is 1mm wide will have roughly 200pH of inductance per millimeter. And again, that is with a ground plane right next to the trace, helping to cancel out most of the magnetic field and inductance. Even in that case, a 5cm/2inch trace will have enough inductance to yield the rise times you're seeing.
Now look at your setup, and think about any areas anywhere in the signal path, like the internal path in the relay itself, where the signal and return paths are probably further apart and thus will create a more substantial magnetic field and have more inductance.
My point is that 10nH is a very small amount and unless there is some obvious low-hanging fruit in your signal path that is likely contributing a good portion of this parasitic inductance that you can eliminate, be prepared for a potentially costly up-hill battle to reduce the inductance enough to meet your requirements.
3. But what about the fall time?
I know 700ps is within spec, but everything above was really to provide context for the longer fall time.
When you have energy stored in a magnetic field, it doesn't just disappear. If you abruptly stop or interrupt the current that had been generating those magnetic fields, those fields collapse and induce current to flow in the conductor (due to plain old Faraday's law of induction - collapsing magnetic fields are definitely changing magnetic fields, and thus they induce a voltage). Unfortunately, the voltage these collapsing fields induce will far exceed 250V. You've tried to very rapidly interrupt the current flow, so the voltage (also called back EMF) will rise as high as is necessary. The energy has to go somewhere, and the only other option is an electric field. Storing energy in an electric field requires capacitance.
So guess what the recently opened contacts of your relay are now providing? Yep, capacitance. The contacts are two plates separated by a dielectric. A 3ps break time for a 50ns pulse into 50Ω at 250V could produce a back EMF pulse as high as ~27kV depending on conditions.
Regardless, the voltage will be plenty high enough to arc across the initially small gap between the relay contacts, and once that ion channel is formed, it doesn't need a high voltage to sustain it. Current keeps flowing even after the contacts have opened, which means there is not just the initial energy stored from the pulse rise time to deal with, but also on-going energy from the current still flowing from the 250V through the relatively low-resistance arc that has formed between the contacts.
Do note that this voltage spike will be a voltage across the relay contacts, meaning it is also dropping across them. So even while all this is occurring inside the relay, the spike will not show up in your pulse at the other end of the transmission line, but will be across the switch itself only. So even if you don't see some awful kV transient on your 50Ω load, trust me, it's there. But only across the relay.
This is why relays will have different make and break ratings, because the act of breaking a flowing current compared to making contact when there is obviously no current flowing are not symmetric at all.
4. Conclusion
The answer here is, unfortunately, probably not the one you had hoped to hear. The longer fall time compared to the rise time is both normal and expected. And there isn't much you can do about it except figure out how to reduce your parasitic inductance.
The mercury-wetted relay is a hundred times faster than your rise times and your rise and fall times are being caused by line inductance (or parasitic inductance due to the relay geometry itself, or likely small contributions of many different parasitic sources throughout your signal path) and you'll need to lower that inductance to get faster rise and fall times.