Are any two complexity classes equipped with an oracle to solve the halting problem equivalent?
No, they are not in general - and focusing on the halting problem specifically makes the situation look more complicated than it actually is. The real point is the following:

There are lots of separation results which relativize to all oracles.

There's a slight technical issue here, actually: how exactly should we treat space-based complexity classes in the oracle context? See the bottom of page $5$ here. I'll only talk about time-based classes here to avoid having to think about this.

For example, the classical argument that $\mathsf{P}\subsetneq\mathsf{EXPTIME}$ in fact shows that $$\mathsf{P}^A\subsetneq\mathsf{EXPTIME}^A$$ for every oracle $A$. In particular, we can take $A$ to be the Halting Problem. Another example of a separation which relativizes to all oracles - including the halting problem - is the time hierarchy theorem. 

This topic is treated in chapter $3$ of Arora/Barak: 


  "Any result about TMs or complexity classes that uses only Properties I and II above also holds for the set of all TMs with oracle $A$. Such results are called relativizing results.
  All of the results on universal Turing machines and the diagonalizations results in this chapter are of this type." 
  
  (pg. 72; I've replaced their "$O$" with an "$A$" to avoid conflict with your use of "$O$" for the halting problem specifically.)


Of course not everything relativizes - and see in particular Baker-Gill-Solovay - but some things do relativize, and all relativizing separation results yield negative answers to the OP as special cases.



Just for completeness, here's one proof that $\mathsf{P}^A\not=\mathsf{EXPTIME}^A$ for any oracle $A$ goes; unsurprisingly, it's a direct diagonalization. Fix an oracle $A$; we want to show $\mathsf{P}^A\subsetneq\mathsf{EXPTIME}^A$. We'll use an exponential-time machine with oracle $A$ to diagonalize against polytime machines with oracle $A$. Fix appropriate enumerations $(\Phi_i)_{j\in\mathbb{N}}$ and $(p_j)_{j\in\mathbb{N}}$ of oracle machines and polynomials respectively and an appropriate pairing function $\langle\cdot,\cdot\rangle$, and let $$X=\{\langle a,b\rangle: \neg\Phi_a^A(\langle a,b\rangle)[p_b(\lfloor\log (\langle a,b\rangle)\rfloor)]\downarrow=1\}$$ (here "$[s]\downarrow=$" means "halts in $s$ steps and equals"). By construction we have that $X\not\in \mathsf{P}^A$ (suppose $\Phi_a^A$ decided it in time $p_b$ ...); however, the "brute force" calculation of $X$ from $A$ only takes exponential time, so $X\in \mathsf{EXPTIME}^A$.

Note that this is just the usual proof, with an "$A$" superscript added everywhere. The point is that diagonalization is such a "coarse" technique that - when it works at all - it tends to work for all objects like Turing machines. And this includes Turing machines relative to a fixed oracle.