Yes. The instantaneous frequency, multiplied by the square modulus of the transformed function, upon integration, yields a Parseval theorem result for the original signal - provided that it has been first reduced to an analytic signal. If it has not been, then the result is integral of the ± square modulus of the signal, in the frequency domain, with a + sign for positive frequency components and a - sign for negative frequency components.
For the wavelet transform $\tilde f(τ,σ)$, of a function $f(t)$, the derivative of $(\arg \tilde f(τ,σ))$ with respect to $τ$ is $2π$ times the instantaneous frequency at $ν_f(τ,σ)$ at $(τ,σ)$. The integral of $ν_f(τ,σ) |\tilde f(τ,σ)|^2$, up to a normalization factor, is the desired result.
For this result, it does not matter what the inverse transform is - or if there even is one. So, it applies to any transform whose forward direction is a wavelet transform. In particular, it applies to both the S and Q transforms; except that the Q transform is normally only defined discretely, not as a continuous transform, though it can be, in which case it becomes a generalization of the S transform.
In particular, if $f(t)$ is the signal and its Fourier transform is defined by
$$\hat f(ν) = ∫ f(t) 1^{-νt} dt$$
where - for convenience - I am writing complex sinusoidals as $e^{2πix} = 1^x$, and if the wavelet transform is written as
$$\tilde f(τ,p) = ∫ f(t) \overline{ψ(p(t - τ)) |p|^A} dt$$
where - for convenience - I'm using $p = 1/σ$ in place of $σ$ (it plays a role analogous to the frequency in the time-frequency plane), then
$$∫ ∫ |\tilde f(τ,p)|^2 ν_f(τ,p) {dτ dp \over |p|^{2A}} = \left(∫ \left|\hat f(ν)\right|^2 sign(ν) dν\right) \left(∫ |\hat ψ(P)|^2 dP\right).$$
The choice $A = ½$ makes this an integral over the log scale in $p$ (or in $σ$), with the integral measure $dτ dp/|p|$ or $dτ dσ/|σ|$, while $A = 0$ makes it an integral with over $p$ as a linear scale with measure $dτ dp$, and $A = 1$ makes it an integral with measure $dτ dσ$, with $σ$ now being on a linear scale.
This requires that
$$0 < ∫ |\hat ψ(P)|^2 dP < ∞$$
For the wavelet transform, since the same windowing function is used in both the forward and reverse directions (with $A = ½$), there is already another admissibility condition
$$0 < ∫ |\hat ψ(P)|^2 {dP \over |P|} < ∞$$
so that it can be normalized. So, the integral admissibility condition has to hold in addition to the transform's admissibility condition.
A much weaker admissibility condition is required for the S transform that can, for instance, be satisfied by enveloped complex sinusoidals; so, it is easier to do.
Derivation:
The actual expression for the integrand could be written in a form evocative of what's often seen in the literature in Quantum Field Theory
$$|\tilde f(τ,p)|^2 ν_f(τ,p) = \overline{\tilde f(τ,p)} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) \tilde f(τ,p),$$
where $\overleftrightarrow{∂/∂τ}$ is $∂/∂τ$ applied to the right minus $∂/∂τ$ applied to the left. (Edit: I previously wrote, by mistake, that it was half right minus half left. That convention is also in use in the Physics literature.) In particular,
$$∫ \overline{1^{ν'τ}} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) 1^{ντ} dτ = ∫ 1^{-ν'τ} \left(ν + ν' \over 2\right) 1^{ντ} dτ = {ν + ν' \over 2} δ(ν - ν') = ν δ(ν - ν').$$
The wavelet transform can be written in the frequency domain as
$$\tilde f(τ,p) = ∫ \hat f(ν) \overline{\hat ψ\left({ν \over p}\right)} 1^{ντ} |p|^{A-1} dν.$$
Throwing this all together, we get
$$\begin{align}
∫ \overline{\tilde f(τ,p)} \left({1 \over {4πi}} \overleftrightarrow{∂ \over ∂τ}\right) \tilde f(τ,p) dτ
& = ∫ ∫ \overline{\hat f(ν')} \hat ψ\left(ν' \over p\right) |p|^{2A-2} ν δ(ν - ν') \hat f(ν) \overline{\hat ψ\left(ν \over p\right)} dν' dν \\
& = ∫ {\left|\hat f(ν)\right|}^2 {\left|\hat ψ\left(ν \over p\right)\right|}^2 ν |p|^{2A-2} dν.
\end{align}$$
Applying this to the integration with $dp/|p|^{2A}$, we have
$$\begin{align}
∫ ∫ |\tilde f(τ,p)|^2 ν_f(τ,p) {dτ dp \over |p|^{2A}} & = ∫ ∫ {\left|\hat f(ν)\right|}^2 {\left|\hat ψ\left(ν \over p\right)\right|}^2 {ν dν dp \over |p|^2} \\
& = ∫ ∫ {\left|\hat f(ν)\right|}^2 sign(ν) {\left|\hat ψ\left(ν \over p\right)\right|}^2 dν {|ν| dp \over |p|^2},
\end{align}$$
and finally, with the substitution $P = ν/p$ and the separation of the $ν$ and $P$ integrals out from one another, we have result originally cited above.