I have not found a proof of precisely this identity, but it follows quickly from the equations in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. The Fourier cosine and sine transforms of $\phi_m$ are equal to the real and imaginary parts of $(1-z)z^m$, with $z=(2k-i)/(2k+i)$. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=-\frac{2}{\pi}\frac{(2 m+1) (\sin \pi m \sin \pi n-1)+(2 n+1) \cos \pi m \cos \pi n}{2 (m-n) (m+n+1)}$$
which is a different way to write the answer in the OP.