No, homogeneity is not required when applying this theorem to monotone circuits, and in fact homogeneity is a quite technical restriction that can be removed even in the general case by weakening the theorem a bit. I'll try to explain some things that made this more clear to me. (and hopefully being wrong in the internet will be more provocative than asking a question)
For completeness, the definition of monotonicity is that the subset of field elements used by the circuit cannot produce $0$ by additions and multiplications. The definition of homogeneity is that all gates of the circuit compute a homogeneus polynomial, i.e., a polynomial whose all monomials have the same degree.
The reason why homogeneity is quite technical restriction is that if $f$ is a polynomial that can be computed by an arithmetic circuit of size $t$, then the polynomial $f_d$ consisting of degree $d$ monomials of $f$ can be computed by a homogeneous arithmetic circuit of size $O(d^2 t)$. (This is for example stated as Lemma 2 in the Hyafil's paper.) This means that we can remove the homogeneity requirement from the theorem if we allow a blow up of a factor $O(d^3)$ in the number of terms.
Now, let's see what we need to assume so that the proof works:
First, given a circuit computing a polynomial $f$ with degree $deg(f)$, we need to find a gate $v$ computing a polynomial $f_v$ with degree $deg(f)/3 \le deg(f_v) \le 2deg(f)/3$. We need to assume fan-in 2 for this, but no other assumptions are required, as for any polynomials $g,h$ it holds that $deg(g) + deg(h) \ge deg(g \diamond h)$, where $\diamond$ is either the sum or the product.
Now that we have found the gate $v$, we write the polynomial $f$ as $f = f_v \cdot h + f_{v=0}$, where $f_{v=0}$ is the polynomial obtained from the circuit by setting $v$ to zero. No assumptions are required to see that the polynomial $f_{v=0}$ can be computed with one gate less than $f$, and any assumptions we made about the original circuit that are closed under setting a gate to zero also hold for the circuit computing $f_{v=0}$ (in particular, monotonicity or homogeneity).
One more important thing is that we have to guarantee that the degree of $f_{v=0}$ is at most $deg(f)$.
This holds if we cannot remove $v$ from the circuit without affecting the output and if for all gates in the circuit it holds that $deg(g \diamond h) \ge \max(deg(g), deg(h))$, where $\diamond$ is the operation computed by the gate and $g$ and $h$ the inputs.
This inequality holds for monotone polynomials and for homogeneous polynomials whenever $g \diamond h$ is non-zero, and here we indeed need to assume either monotonicity or homogeneity in order to make the proof work.
What is left is to show that there is such polynomial $h$ with $deg(h) \le 2deg(f)/3$.
The existence of $h$ follows by the observation that the polynomial $f_{v=0}$ is the same as the polynomial $f$, except that some monomials with a factor $f_v$ are missing. This can be proved by induction by separating the polynomials computed by each gate to two parts, the part coming from $v$ and the other part.
The degree of $h \cdot f_v$ is at most $deg(f)$ because the degrees of $f_{v=0}$ and $f$ are at most $deg(f)$.
Now, $deg(h) \le 2deg(f)/3$ follows from the facts that $deg(f_v) \ge deg(f)/3$ and $deg(h \cdot f_v) = deg(h) + deg(f_v)$, which is because the polynomials are over a field, which is a standard assumption in arithmetic circuits [1].
This would complete the proof of the version of the theorem that I stated in the question, but that version is missing an important property: if we assume either homogeneity or monotonicity, can we assume homogeneity or monotonicity of $f_v$ and $h$? For monotonicity, we see that the process of separating $h$ from $f$ by induction is closed under monotone arithmetic. For homogeneity, we have that $f_v$ and $f_{v=0}$ are homogeneous, so $h$ cannot be non-homogeneous because then $f$ would be non-homogeneous.
To summarize, the assumptions of this theorem are:
- The polynomials are over a field. This is a standard assumption in arithmetic circuits [1].
- The gates of the circuit have fan-in 2.
- The circuit is homogeneous or monotone, or we have some other argument to guarantee that $deg(f + g) = \max(deg(f), deg(g))$ for $+$ gates with inputs $f$ and $g$.
[1] Shpilka, Amir, and Amir Yehudayoff. Arithmetic circuits: A survey of recent results and open questions. Now Publishers Inc, 2010.