What would be the most energy-efficient way (using known physics) to blow apart a star or otherwise prevent or greatly slow the rate at which it performs fusion?
Spin it until fusion stops. Do this using the sun's own energy.
To accomplish this, I will have to ask you to envision something like a Dyson Sphere, but the primary function of the matter encircling the star will be mirrors. I will be making statements about the force balances and basic physics, how this could be actually done in practice is out of the scope of this answer.
I propose that mirrors would at some distance from the sun, and the focal length of these mirrors would be equal to this distance. The idea will be to reflect the sun's light back onto the sun in a way that makes it spin faster. We would like to redirect all the light nearly tangentially to the edge of the sun, but we can't do this because of entropic limits. Remember, nothing can be focus the sun's rays to heat something hotter than the surface of the sun. This is why I select the parameter of $R=f$.
Focal length equal to distance, per Wikipedia
With this type of mirror we could focus the sun directly back onto the sun. We will assume sufficient distance from the sun to treat it as a simple circle. To make the sun spin faster, we will redirect the image right of the center, so the image will have a center a distance of $d$ to the right of the object. In order to get the optimal location to direct the reflection, we will
- Integrate the distance from the axis of rotation from the lower bound of the image circle to the upper bound of the object circle
- Integrate this value between the two intersections of the circles
- Find the greatest value of this moment integral over all valid values of $d$
I actually did that calculation. I obtained $d=0.836 R$. To summarize, the proposal is to refocus the light back onto the sun so that it looks like a Venn diagram.

By doing this, we trash a lot of the radiation. I calculate the average radius of interaction to be $0.418 R$, but if we dilute that number by the number of photons lost, we will get a multiplier of $0.202 R$ to convert the photon's momentum to the average torque exerted. I believe this is a fundamental, entropic limit, and if I'm wrong, it's at least conservative.
At this point we're still not finished. That's because this mirror isn't balanced. If it always reflected light in this way, it would not have a stable orbit since the gravitational force can only act radially and there is a tangential component to the photon's force on the mirror. One could compensate for this quite simply by holding a flat mirror at a $45^{\circ}$ angle to the sun's radiation, directing them in the other tangential direction. The torque from that balancing mirror, however, would depend on the distance from the sun. Because of that, we can't numerically correct for it here. If the distance of this satellite from the sun was large compared to the sun's radius then the loss from this balancing mirror would be negligible.
For simplicity I'll assume all the radiation from the sun is photons (it's only about 98%). The power output of the sun is:
$$P = 3.846 \times 10^{26} W$$
The photonic momentum (which is totally isotropic normally) can be found from $E=pc$ applied to the above power output. This gives the total momentum of photons emitted per unit time, or in other words, the isotropic photonic force.
$$F = P/c = 1.282 \times 10^{18} N$$
The moment of inertia of the sun can be found from the forumula for the moment of inertia for a solid ball.
$$m = 1.989 \times 10^{30} kg$$
$$I = \frac{2}{5} m R^2 = 3.848 \times 10^{47} m^2 kg$$
Using the methods I've laid out here, I can calculate the torque. This is assuming that all photons from the sun are used as efficiently as possible.
$$ \tau = F \bar{R} = (1.282 \times 10^{18} N) ( 0.202 R) = 1.801 \times 10^{26} N m $$
How fast would it need to spin in order to stop fusion, and then how much to break it apart? This is a difficult question to answer. However, one thing we can say is that if you have enough energy to completely disociate everything in the sun gravitationally you have enough energy to do both of the tasks of stopping fusion and breaking it apart less spectacularly. The energy to fully spread out the sun's mass over all space can be calculated. This is similar to a recent question, in short, the full dissociation energy is the half of the potential integrated over the entire volume. If I've done this right, the full disociation energy is:
$$ E_{diss} = \frac{3 G M^2}{8 R} = 1.423 \times 10^{41} J$$
It is also difficult to estimate the time needed for the available torque to dissociate the sun. But let's do a limit case where the sun doesn't deform due to the increased rotation. In that case we can seek an angular velocity that is equivelant to the above energy of dissociation, then ask how long it would take the available torque to accelerate it to that point.
$$ E_{diss} = \frac{1}{2} I \omega^2$$
$$ \omega = 0.00086 \frac{rad}{s}$$
This seems small, but consider that this would be the state of rotating once every 2 hours. Now, how long would the given torque take to get it to this state?
$$ I \omega = \Delta t \tau $$
$$ \Delta t = 58.23 \text{ billion years} $$
Now, if someone started spinning the sun with the sun's own power, eventually the fusion would stop, but the radiation wouldn't stop right away. In order to see if the stored energy is sufficient to break the sun apart, we'll consider the thermal energy stored in the core alone. The core's temperature is about 15,000,000 kelvin. This region goes out to around 0.25 solar radii. The average kinetic energy of a helium nuclei in the sun's core would then come from $E_k = 3/2 k T$, coming out to
$$ E_k = 3.106 \times 10^{-16} J $$
The average molecular mass in the sun is about 1.67
http://web.njit.edu/~gary/321/Lecture7.html
I can use this to find the number of nuclei in the core of the sun. I can then combine that with the previous value for energy per nuclei to find the total stored kinetic energy in the sun.
$$ E = E_k N = E_k m (0.25)^3 / (1.67 amu) = E_k (1.12 \times 10^{55} \text{particles} ) = 3.481 \times 10^{39} J $$
This is the total thermal energy stored in the sun's core. Roughly. Divide this by the normal power output to get a $s$ valued number for the sun's power.
$$ E / P = 9.05 \times 10^{12} s = 6.9 \text{million years} $$
We conclude that the stored energy of the sun is insufficient to fully break it apart by about 4 orders of magnitude. The described method would still be viable to stop the fusion reaction.