_Edit: In revision 1, I wrote an embarrassingly complicated answer.  The answer below is much simpler and stronger than the older answer._

> Let O be a complexity class that is closed under complement, i.e. O=coO. Also we assume that the logspace, L, is a subset of O.  
> […]  
> Do we have any oracle O such that L<sup>O</sup>≠O?

Yes.

First assume NP≠coNP, and let O=NP∪coNP.  Then O is closed under complement and O contains L.  However, note that L<sup>O</sup>=L<sup>NP</sup>, and in particular L<sup>O</sup> has a complete problem <i>X</i> under polynomial-time many-one reducibility.  If L<sup>O</sup> is equal to O=NP∪coNP, then <i>X</i> belongs to either NP or coNP, and in either case, NP=coNP, contradicting our assumption.

Next we remove the assumption NP≠coNP.  Note that the above argument relativizes.  Therefore, take any language <i>A</i> such that NP<sup><i>A</i></sup>≠coNP<sup><i>A</i></sup>, and O=NP<sup><i>A</i></sup>∪coNP<sup><i>A</i></sup> gives an unconditional example such that L<sup>O</sup>≠O.