The RQ-125B sounds like a solid choice, and as you note, the only puzzle you will have to solve is where you will get the +8.5V from.
Looking at the datasheet, two things stand out to me. At +5V, the RQ-125B has a rated current of 11A. At +12V, the rated current is 4.5A.
Your current measurements for the power rails is really valuable --- that's new information. I'm personally somewhat pleased to see that my estimate of the +5V current was pretty accurate
But anyway, note the following:
You are already consuming a measured 2.8A on +12V, and that leaves 4.5A - 2.4A = 2.1A left over in your current budget.
Do you know how much +12V current your +8.5V buck converter would consume when it's serving out 2.4A? It could be less than the 2.1A you have left. But there may be some loads on +12V that you may not have tested yet:
Those tape select magnets are electromagnets that turn on whenever the tape drive wants to move the tape:
I have not measured how much power they use, but I would assume that they use a fair amount of current in order to press the jackshaft rollers against the spindle.
I'm concerned that you may overrun the RQ-125B's +12V current budget of 4.5A!
There may be another option for an all-in-one solution though: instead of getting +8.5V by converting down +12V, you might try upconverting from +5V with a
boost converter. Let's round your measurement for the +5V current, 4.3A, to 5A to add some safety margin: you now have 6A remaining out of the RQ-125B's rated 11A to dedicate to making +8.5V. Power = current*voltage, so 6A * 5V = 30W. Now, 30W / 8.5V = 3.5A: I suspect whilst making no guarantees that as long as you get an efficient boost converter (since it will consume some current of its own), then this will give you the power you need with some margin. This is the approach I would likely prefer, although more study is probably advisable.
More notes in a follow-up post.