There's many ways to answer that, here are some to throw out ideas:
1) since it is a second order filter with 2X slope per octave then you can argue that it should be sqrt(10) X.
2) typically though that measurement might also take into account the ripple and constrain it so that a fixed factor may not be apropos. In your case of critically damped this doesn't hold, but in a more general sense perhaps a simple factor isn't the right criteria.
3) Since I work with S&H system and switched cap we don't use a 10X but how many time periods until settled to within the error budget of the system (so it's variable depending upon what is being designed).
I am afraid, changing the opamp type will not help. The observed effect (less damping for rising frequencies) is the typical disadvantage of the lowpass Sallen-Key topology.
The reason is as follows: For rising frequencies the "classical" output signal from the opamp decreases (as desired) - however, at the same time there is a signal arriving at the output via the feedback capacitor (the signal bypasses the opamp). This signal produces an output voltage across the finite output impedance of the opamp (the output impedance even increases for rising frequencies). Hence, this unwanted signal dominates for high frequencies and limits the damping at a fixed value.
If you need more damping for very large frequencies the only solution is to use another filter topology (Sallen-Key/negative, multi-feedback MFB, GIC,..).
The same effect can be observed for the classical inverting Miller integrator (capacitor in the feedback path).
EDIT/COMMENT: Of course, this unwanted effect can be suppressed using another buffer amplifier within the positive feedback path (driving the feedback capacitor). However, this method requires another opamp.
EDIT2: Depending on your damping requirements - it could be sufficient to use another filter topology (MFB) for the last of the three filter stages only. As another alternative, you could add a passive RC lowpass and and a buffer stage after the third filter stage.
EDIT3: Here is a simple "trick" for improving the attenuation of the existing filter circuit in the stop band: Modify the impedance level of the parts used. For example: Increase all resistors by a factor k (for example: k=10) and reduce all capacitors by the same factor. Thus, all time constants and the whole filter respose remains unchanged, but the direct way to the opamp output now contains a larger resistors (R2, R4, R6) and a smaller capacitor. This should decrease the remaining voltages at the output for very large frequencies to a value of app. **r,out/(r,out+RX)**with RX=R2, R4, R6, respectively.
Best Answer
Forget Sallen/Key for the moment, go back to the equivalent passive implementation of the same filter. (e.g. search "passive butterworth filter" for examples.) You'll see it involves an L and a C per 2nd-order filter section, thus it's basically an L-C resonant circuit, with resistive damping to control the resonance (the height of the knee).
Having understood this, then you can use a C in the feedback path of an amplifier to simulate the L in an L-C resonant circuit - and the Sallen-Key circuit is basically one example of this. As "FakeMoustache" comments, the gyrator is another.