For completely submerged bodies the buoyance force, being simply equal to the weight of the displaced fluid, is stronger for a denser fluid.
But you know that the buoyancy force for a partially submerged body (like a sailing boat) must be equal to the weight of the body (unless the boat sinks or starts flying like a balloon).
Since the buoyant force is equal to the weight of the displaced fluid, a (non-sinking) boat displaces always the same mass, no matter which fluid, but more volume of a less dense fluid.
A classical example happens if you submerge an egg in water. It sinks to the bottom of the top. Then start adding salt, until eventually the egg will raise. See for example Tommy's webpage:
A quite different question is if a boat would happily float in a denser fluid like mercury, without turning upside down. The shape of the submerged part is very important for the stability. The buoyancy centre must be higher than the centre of mass, otherwise it will be unstable (that is why ballast is needed in many cases, to make a boat heavier in its underwater part... too much of the boat above water would result in a dangerous high centre of mass)
EDIT: Ok, when the partially submerged body is in equilibrium, then
$$W_{\text{displaced fluid}}=W_{\text{object}}$$
$$\rho g \Delta V = W_{object}$$
Since $g$ and the weight of the object $W_{\text{object}}$ are fixed, an increase in density means a decrease in the submerged volume, for the equation to hold.
The density of water is 1 g/ml. If the volume of the object is 100 ml and the object weighs 90 g, it is less dense than the water (0.9 g/ml) and hence only 90 ml of the object will be submerged (depending in the shape), leaving 10 ml of the object exposed above the water level.
Best Answer
Let us suppose we remove the object and fill the space left (in the fluid) with the same fluid. Assume this portion of fluid become solid without changing its volume or density. It will be in equilibrium with the fluid. Now suppose the buoyancy force on this solidified portion is off the center of mass. This would imply non vanishing resultant force or torque and the solidified portion would not be in equilibrium. A kind of inverse argument can also be given. Suppose the center of buoyancy coincides with the center of mass of an object immersed in a fluid. Then you would never observe resultant torque (by means of rotations) on any object. And that is empirically not true.