Analysis

This problem examines the odd and even properties of functions and their extremum. To solve this problem, we can determine the analytic expression of the function based on its odd and even properties, then use the properties of the hook function to find the extremum, and finally solve the inequality that always holds.

Solution

Given the problem, when $x > 0$, $f(x)=-f(-x)=-(-9x- \frac{a^2}{x}+7)=9x+ \frac{a^2}{x}-7$.

Based on $f(x) \geqslant a+1$ holding for any $x \geqslant 0$,

Therefore, when $x=0$, $f(0)=0 \geqslant a+1 \Rightarrow a \leqslant -1$,

When $x > 0$, $f(x)=9x+ \frac{a^2}{x}-7 \geqslant a+1$ always holds, i.e., $f_{\min}(x) \geqslant a+1$.

The function, according to the properties of the hook function, has $f(x)=9x+ \frac{a^2}{x}-7$ reaching its minimum value when $9x= \frac{a^2}{x} \Rightarrow x=| \frac{a}{3}|$, thus $f_{\min}(x)=6|a|-7$,

Therefore, $6|a|-7 \geqslant a+1$,

Since $a \leqslant -1$, we get $-6a-7 \geqslant a+1 \Rightarrow a \leqslant - \frac{8}{7}$.

Hence, the range is $\boxed{(-\infty, -\frac{8}{7}]}$.