Since the angle between the two unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $120^{\circ}$, we have $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| \cdot |\overrightarrow{b}| \cos 120^{\circ} = -\dfrac{1}{2}$.

Therefore, $|\overrightarrow{a}-k\overrightarrow{b}|^{2} = \overrightarrow{a}^{2} - 2k\overrightarrow{a} \cdot \overrightarrow{b} + k^{2}\overrightarrow{b}^{2}$

$= 1 + k + k^{2} = (k + \dfrac{1}{2})^{2} + \dfrac{3}{4} \geqslant \dfrac{3}{4}$.

It follows that when $k = -\dfrac{1}{2}$, the minimum value of $|\overrightarrow{a}-k\overrightarrow{b}|$ is $\dfrac{\sqrt{3}}{2}$.

Hence, the correct choice is $\boxed{B}$.

This solution utilizes the definition and properties of the dot product of vectors: the square of a vector equals the square of its magnitude, combined with the method of finding the minimum value of a quadratic function. This question tests the understanding of the definition and properties of the dot product of vectors, specifically that the square of a vector equals the square of its magnitude, and the method of finding the minimum value of a quadratic function, assessing computational skills. It is considered a medium-level question.