For an object that is fully submerged the buoyant force is constant (in both magnitude and direction) exactly as the usual "near the surface of the Earth" approximation to gravity. That part is clearly conservative.
To figure the behavior of a object while partly submerged we much rely on Archimedes' principle, which tells us that the buoyant force is proportional to the weight of the volume then displaced. The act of submerging the body may require a different distance depending on the shape and orientation of the body, so we can be suspicious.
Take for a concrete example a non-spherical, ellipsoidal object. By symmetry and construction it will require the same average force to insert it into the fluid at any angle, but that force will be applied over a different distance depending on orientation. The act of submerging the object itself requires different amounts of work depending on the orientation of the object.
But before we say "Ah! Ha!" we need to consider that the starting and ending conditions for putting it end the long way are separated by a larger total distance than those for putting it in the short way. To know if the force is conservative over those paths we have to make the starting and ending points the same. Allow the starting point of the COM to be a distance $a$ over the fluid surface and the ending point to be a distance $a$ under the surface (where $a$ is the semi-major axis of the ellipse in some cross-section). The work done by gravity plus buoyancy on a vertical path between those points is
$$W_a = 2a\left(\rho_o - \frac{1}{2} \rho_f \right)V_o g\;,$$
where $V_o$ is the volume of the object and the densities $\rho_{o,f}$ are for the object and the fluid respectively. Compare to lowering the object oriented so that the minor axis $2b$ is vertical. The work done is
\begin{align*}
W_b
&= \left[(a-b) \rho_o V_o g\right] + \left[2b\left(\rho_o - \frac{1}{2} \rho_f\right)V_o g \right]+ \left[(a-b) \left(\rho_o - \rho_f\right)g V_o\right] \\
&= \left[ \left((a-b) \rho_o\right) + \left(2b\left(\rho_o - \frac{1}{2} \rho_f\right) \right)+ \left((a-b) \left(\rho_o - \rho_f\right)\right)\right]V_o g \\
&= \left[ \left[ (a-b) + 2b + (a-b) \right] \rho_o - \left(b + (a-b)\right)\rho_f\right]V_o g \\
&= \left[ 2a\rho_o - a\rho_f\right]V_o g \\
&= 2a\left( \rho_o - \frac{1}{2}\rho_f\right)V_o g \;,
\end{align*}
exactly as before.
This shows that at least for sufficiently symmetric objects and considering the combination of gravity and buoyancy even the dunking process is conservative. Explicit extension to arbitrary shapes requires that the summation done here be performed as an integral with careful attention to the endpoints and the completely out of water and completely underwater parts of the movement, but one might also appeal to intuition about having to move the same amount of fluid upward to achieve submersion.
