The answer is yes, in a very general way.

What I claim, first, is that the Lévy-Montague reflection theorem holds in ZF for any definable continuous cumulative hierarchical representation of the set-theoretic universe $V$. That is, if you have defined sets $U_\alpha$ for every ordinal $\alpha$, such that

- the sequence $\alpha\mapsto U_\alpha$ is definable, without parameters
- the sequence is monotone $\alpha\leq\beta\implies U_\alpha\subseteq U_\beta$
- the sequence is continuous, $U_\lambda=\bigcup_{\alpha<\lambda}U_\alpha$ for limit ordinals $\lambda$
- the sequence accumulates to the entire universe, $V=\bigcup_\alpha U_\alpha$

Then for every formula $\varphi(x)$ in the first-order language of set theory, there is some ordinal $\alpha$ such that $\varphi$ is absolute between $U_\alpha$ and $V$. Indeed, there is a closed unbounded class of such ordinals.

The proof is nearly identical to the usual proof of the reflection theorem — these hypotheses are what is used about the $V_\alpha$ hierarchy.

Once one has reflection, then one can define HOD using the hierarchy, as you suggest, by saying a set $a$ is *ordinal definable*, if there is some $\alpha$ and some ordinal parameters below $\alpha$, such that $a$ is definable in $U_\alpha$ from those parameters. A set is *hereditarily ordinal definable, if $a$ and every hereditary member of $a$ is ordinal definable.

In any model $V$ of ZF, if a set is ordinal definable, then by reflection it is ordinal definable in that sense. And conversely, if a set is ordinal definable in that sense, in some $U_\alpha$ using some formula $\varphi$ with ordinal parameters $\vec \beta$, then in $V$ we may define the set using parameters $\langle\alpha, \vec\beta,\varphi\rangle$ as parameters. (And note the very subtle point that we must sometimes use the formula $\varphi$ itself as a parameter, meaning its Gödel code, since in an $\omega$-nonstandard model the formula $\varphi$ used to define $a$ in $U_\alpha$ might be nonstandad. I view it as a kind of lucky miracle that ordinal-definability doesn't stumble on this problem.)

So this version aligns with the usual version of HOD.

**Conclusion.** It doesn't matter which definable continuous hierarchy you use when defining HOD. They all give the same class.

Another way to see this is to observe the following:

**Theorem.** For any such hierarchy $U_\alpha$ as above, there is a closed unbounded class of ordinals $\theta$ for which $$U_\theta=V_\theta.$$ In particular, any two such hierarchies agree on a class club.

**Proof.** This is the typical club argument. First, by continuity, the class of such ordinals $\theta$ is closed. And it is unbounded, since we can start in any $U_{\alpha_0}$, find a $V_{\alpha_1}$ containing all those elements, and then a $U_{\alpha_2}$ containing all those elements, and so on, building an alternating chain
$$U_{\alpha_0}\subseteq V_{\alpha_1}\subseteq U_{\alpha_2}\subseteq\cdots$$
If $\theta=\sup_n\alpha_n$ is the supremum, then $U_\theta=V_\theta$ by continuity. So this is a closed unbounded class. $\Box$

Since reflection also works on a class club, we see in this way that every formula $\varphi$ reflects to a ordinal on which the two hierarchies agree.