From $f(0)=3$, we obtain: $a=3$.

From $f(2-x)=f(x)$, we get: $x= \dfrac {b}{2}=1$, solving for $b$ gives: $b=2$.

Thus, $f(x)=x^{2}-2x+3$. $f(x)$ is decreasing on $(-∞,1)$ and increasing on $(1,+∞)$.

For $b^{x}=2^{x}$ and $a^{x}=3^{x}$,

When $x f(2^{x})$, i.e., $f(a^{x}) > f(b^{x})$.

When $x \geqslant 0$, $1 \leqslant 2^{x} \leqslant 3^{x}$, thus $f(2^{x}) \leqslant f(3^{x})$, i.e., $f(b^{x}) \leqslant f(a^{x})$.

In summary, $f(b^{x}) \leqslant f(a^{x})$.

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

By using $f(0)=3$ and $f(2-x)=f(x)$, we can find the values of $a$ and $b$. Then, we can determine the conclusion based on the monotonicity of the function. This problem examines the monotonicity of functions and the properties of quadratic functions, making it a moderately difficult question.