No, this would not be a feasible weapon.
Although I originally posted an answer to this question five years ago, I've since deleted it and am starting anew. One reason I deleted it is that the answer only tangentially addressed the question, despite receiving some upvotes. The other reason is that the portion that does, in my opinion, adequately work towards a proper answer takes the wrong approach. I tried to assess the feasibility of such a weapon by discussing the power radiated by two merging stellar-mass black holes; instead, I should have chosen a quantity that is distinctly characteristic of gravitational waves: strain.
The strain, $h$, of a gravitational wave is a measure of how much spacetime is distorted by the passing wave. If you have an object of length $L$, it will be alternatingly stretched and compressed by a distance $\Delta L=hL$. This is the principle behind the interferometric techniques used by LIGO to observe gravitational waves in the first place.
Say you're a distance $r$ from a pair of binary black holes of total mass $M$. During any point in their coalescence, you can assign them an effective velocity, $v/c$, which increases as the black holes move closer and closer together. The peak strain you measure is then roughly
$$h\sim\frac{GM}{c^2}\frac{1}{r}\left(\frac{v}{c}\right)^2$$
Consider the case of the first detection of gravitational waves. That merger involved two black holes of about $30M_{\odot}$ apiece (and therefore a total mass $M\approx60M_{\odot}$), lying a distance $r=410\text{ Mpc}$ away. Shortly before coalescence, their effective velocity was $v/c\approx0.6$. Plug in the numbers and you find that the strain at that moment, as measured on Earth, should be $h\sim10^{-21}$, which is what was observed by LIGO. That's tiny! Now let's say we're standing a few light-years away from the source - say, $r=1.34\text{ pc}$, the distance from Earth to Alpha Centauri. Now we reach a much larger strain, $h\sim10^{-14}$ - still a pittance. Even at a distance of 1 AU, we only reach $h\sim10^{-7}$. For your weapon to produce observable changes, you'd need to place it in the enemy star system, which I'm reasonably certain would violate the terms of the treaty!
The alternative, of course, is to use much larger black holes. Unfortunately, because the strain is only linearly proportional to the mass of the black holes, you'd need a supermassive black hole binary ($M\sim10^6M_{\odot}$-$10^9M_{\odot}$) to produce noticeable changes at distances of a parsec or so, pushing the existing observational and theoretical limits of the supermassive black holes we know. Moreover, these monsters and the accretion disks and fast-moving clouds around them would constitute significant weapons on their own, and I'd bet that it would be extremely hard to control them - even for a civilization of the capacity you describe.
Miscellaneous notes:
The power of the Hawking radiation from each black hole is about $L\sim10^{-31}\text{ W}$, so it is far from dangerous. Furthermore, they are extremely cool and therefore will only be able to produce extremely low-energy photons, posing no threat whatsoever to any civilization in that sense.
The gravitational wave emission will not be isotropic, as you might imagine, given that the binary is not spherically symmetric. Therefore, there are ways you could angle the system such that your target receives the maximum emission and you receive the minimum (but, as per our analysis earlier, you're in no danger provided you stay far enough away).
The effective range of the black holes is much less than 1 AU; for example, at that distance, the maximum strain is $h\sim10^{-7}$, optimistically.
Would it violate the existing terms of the treaty? As the Hawking radiation would be negligible, you're safe on that front; given that you'd have to enter the other system for the weapon to have any effect, I'd argue that attempting to use this as a weapon would indeed violate the clause regarding "gravitational disruption".
You can show that near coalescence, $v/c\approx0.7$. This is because you can show from Kepler's third law that $v/c\approx R_S^{1/2}/R^{1/2}$, where $R$ is the distance between the centers of the black holes and $R_S$ is the Schwarzschild radius of one of the black holes, assuming approximately equal mass. Just before they meet, $R=2R_S$, so $v/c\approx\sqrt{(1/2)}\approx0.71$.
