There are at least two problems with this expression.
1. The integrand of the far left expression is always zero; hence, even if the expression were true, it wouldn't tell us much.
2. There's a problem with what you're integrating over. The divergence theorem really says that for an enclosed volume, the volume integral equals the given surface integral over the enclosing volume. I could be wrong, you topologists out there could probably answer me yea or nay, but if a volume has an enclosing surface, then I don't think it's possible for that enclosing surface to have an enclosing contour, which is what the Kelvin-Stokes theorem requires. That is, the Kelvin-Stokes theorem would, I think, require a surface to be integrated over which has a boundary. But if it has a boundary, I don't think it can enclose a volume.