Improper Integral - Odd Function and Symmetric range but Diverge

No there is not.

To check if the integral of the function converges, you need to check that

$$\int_0^\infty \vert f\vert <+\infty.$$

Since $f$ is odd, if this value is indeed finite, you also have

$$\int_{-\infty}^0 \vert f\vert=\int_0^{+\infty} \vert f(-x)\vert\mathrm dx=\int_0^\infty \vert f\vert <+\infty.$$

So such a counter-example does not exist, since

$$\int_{-\infty}^{+\infty} \vert f\vert=2\int_0^{\infty} \vert f\vert <\infty.$$

No. You are assuming that $f$ is odd and that the limit $\lim_{R\to+\infty}\int_0^Rf(x)\,dx$ exists. But then$$\int_{-\infty}^{+\infty}f(x)\,dx=\lim_{R\to+\infty}\int_{-R}^Rf(x)\,dx=\lim_{R\to+\infty}0=0.$$The second equality comes from the fact that $f$ is odd.