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^O≠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^O=L^NP, and in particular L^O has a complete problem X under polynomial-time many-one reducibility. If L^O is equal to O=NP∪coNP, then X 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 A such that NP^A≠coNP^A, and O=NP^A∪coNP^A gives an unconditional example such that L^O≠O.

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

Edit: Even the “simplified” answer in revision 2 was more complicated than necessary.

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^O≠O?

Yes. Let O=RE∪coRE, where RE is the class of recursively enumerable languages. Then O is closed under complement and O contains L. However, note that L^O=L^RE, and in particular L^O has a complete language under polynomial-time many-one reducibility. On the other hand, O=RE∪coRE does not have a complete language under polynomial-time reducibility because RE≠coRE. Therefore, L^O≠O.

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^O≠O?

I guess that there is a simpler and/or stronger answer, but anyway: yes, unless the polynomial hierarchy collapses to the second level (Σ₂P=Π₂P). Assuming Σ₂P≠Π₂P, we have L^O≠O for O=P^Σ₂P[1] (that is, the class of problems solvable in polynomial time by using the Σ₂P oracle only once).

Indeed, O contains L and O is closed under complement. Lemma 3.3 of Buhrman and Fortnow [BF99] shows that Σ₂P △ NP := {(X∖Y)∪(Y∖X): X∈Σ₂P, Y∈NP} is not contained in O unless Σ₂P=Π₂P. Note that Σ₂P △ NP is contained in L^O.

[BF99] Harry Buhrman and Lance Fortnow. Two queries. Journal of Computer and System Sciences, 59(2):182–194, Oct. 1999. http://dx.doi.org/10.1006/jcss.1999.1647

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^O≠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^O=L^NP, and in particular L^O has a complete problem X under polynomial-time many-one reducibility. If L^O is equal to O=NP∪coNP, then X 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 A such that NP^A≠coNP^A, and O=NP^A∪coNP^A gives an unconditional example such that L^O≠O.