For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be:

v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997.

So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration.

Also, suppose you accelerated for a second increment equal to the first; you would get

v_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991.

And one more increment:

v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973.

As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.

Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.