Since $0 < a$, $b$, $c < 1$ satisfy the condition $ab+bc+ac=1$,

it follows that $(a+b+c)^{2} \geqslant 3(ab+ac+bc)=3$

Therefore, $a+b+c \geqslant \sqrt {3}$,

Since $( \dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c})(1-a+1-b+1-c) \geqslant (1+1+1)^{2}$

Therefore, $\dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c} \geqslant \dfrac {9}{3-(a+b+c)} \geqslant \dfrac {9+3 \sqrt {3}}{2}$.

The minimum value of $\dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c}$ is $\dfrac {9+3 \sqrt {3}}{2}$ if and only if $a=b=c= \dfrac { \sqrt {3}}{3}$.

Therefore, the correct answer is $\boxed{D}$.

To determine $a+b+c \geqslant \sqrt {3}$, use the Cauchy-Schwarz inequality $( \dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c})(1-a+1-b+1-c) \geqslant (1+1+1)^2$, which allows us to find the minimum value of $\dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c}$.

This problem examines the minimum value of $\dfrac {1}{1-a}+ \dfrac {1}{1-b}+ \dfrac {1}{1-c}$ and the application of the Cauchy-Schwarz inequality, making it a medium-level question.