After thinking for a while i came up with my own proof. I hope it is helpful to you and correct!
Definition: Frame-Structure
The frame structure used in the following is one of one-periodic signals, with a constant header and a uniform distributed payload. The ratio between payload and total frame is given as $$ p_{\%} = \frac{|payload|}{|frame|} $$
the ratio of samples in the header vs samples per frame.
Example:
$$ |HHH|012|HHH|223|HHH|978|...$$
,here: $p_{\%}=0.5$ .
Proof:
The complex envelope of an FSK-signal is
\begin{array}{cc}
x\left(t\right)=e^{i\left[2\pi\Delta FT\left(\varphi\left(j,T\right)+\left(\frac{t}{T}-j\right)d\left(j\right)\right)\right]} & ,\text{}jT\leq t\leq\left(j+1\right)T\end{array}
with
$$\varphi\left(j,T\right):=2\pi\Delta FT\sum_{l=0}^{j-1}d\left(l\right)+\varphi_{0}$$
and the modulation index $$\eta:=2\Delta FT.$$
We introduce the model by having a look at the ACF of truely periodic, i.e. $p_{\%}\left(x\right)=1 $, discrete signals. In a second step we generalize to partly periodic signals with $p_{\%}\left(x\right)<1$. Let $x$ be a digital FSK signal of length N+1, that is, $x$ is a sampled version of the complex envelope above. Then the ACF of x is defined by
\begin{array}{ccc}
\rho\left[x\right]\left(\tau\right) & = & \sum_{k=0}^{N-\tau}x\left(k\right)x^{*}\left(k+\tau\right)\\
& = & \sum_{k=0}^{N-\tau}e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}\right)d\left(j_{k}\right)+\varphi\left(j_{k},T\right)\right)\right]}e^{-i\left[\pi\eta\left(\left(\frac{k+\tau}{T}-\left(j_{k+\tau}\right)\right)d\left(j_{k+\tau}\right)+\varphi\left(j_{k+\tau},T\right)\right)\right]}
\end{array}
Let $x$ be $\tau_{0}$-periodic, consequently $\tau_{0}$ is an integer-multiple of the symbol duration T measured in samples. Given our data model, we can expect to see a distinct peak in the ACF feature of such a signal. Let $j_{k}:=\left\lfloor \frac{k}{T}\right\rfloor$ be the index of the symbol being modulated at point of time $k\in\mathbb{N}$. While the signal $x$ is $\tau_{0}$-periodic, the series of symbols is $\frac{\tau_{0}}{T}$-periodic, i.e. $j_{k+\tau_{0}}=\left\lfloor \frac{k+\tau_{0}}{T}\right\rfloor =j_{k}+\frac{\tau_{0}}{T}$. Therefore the symbols of a signal are repeated periodically, i.e. $d\left(j_{k+\tau_{0}}\right)=d\left(j_{k}+\frac{\tau_{0}}{T}\right)=d\left(j_{k}\right)$. Hence we can simplify the ACF of x at lag $\tau_{0}$ to
\begin{array}{cc}
\rho\left[x\right]\left(\tau_{0}\right)\\
= & \sum_{k=0}^{N-\tau_{0}}e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}-\frac{k+\tau_{0}}{T}+j_{k}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{i\left[\pi\eta\left(\sum_{l=0}^{j_{k}-1}d\left(l\right)-\sum_{l=0}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)\right]}\\
= & \sum_{k=0}^{N-\tau_{0}}e^{i\left[\pi\eta\left(\left(-\frac{\tau_{0}}{T}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\\
= & \sum_{k=0}^{N-\tau_{0}}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}.
\end{array}
The value $\rho\left[x\right]\left(\tau_{0}\right)$ is decisively influenced by the term $p_{m}:=-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)$. It contains exactly one period of the cyclic symbol stream and it shifts cyclically with time and, thus $p_{m}$ is constant. Accordingly we can simplify the ACF feature for periodical signals to
$$\left|\rho\left[x\right]\left(\tau_{0}\right)\right|=\left|\sum_{k=0}^{N-\tau_{0}}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right|=\left(N-\tau_{0}+1\right).$$
The complex value of the ACF of a truely periodic signal is independent of modulation index and symbol stream.
In the second step we assume a partly-periodic signal x of ratio $p_{\%}$ with lag $\tau_{0}$. The ACF of a uniformly distributed series of samples converges to zero at all non-zero lags. Thus, according to our model the autocorrelation of the data part of the frames all but vanishes. To simplify the analysis, we introduce an indicator function b; indicating wether a sample is part of the periodic share of a signal:
$$\begin{array}{ccc}
b\left(k\right) & := & \begin{cases}
1 & k\in\text{header-part of x}\\
0 & \text{else}
\end{cases}\end{array}$$
We get
\begin{array}{cc}
\rho\left[x\right]\left(\tau_{0}\right)\\
= & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}-\frac{k+\tau_{0}}{T}+j_{k}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{i\left[\pi\eta\left(\sum_{l=0}^{j_{k}-1}d\left(l\right)-\sum_{l=0}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)\right]}\\
= & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{i\left[\pi\eta\left(\left(-\frac{\tau_{0}}{T}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\\
= & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}.
\end{array}
The term $p_{m}$ (see [eq:ACF_FSK_PhaseMemory]) is not constant like it was above. As the frames are not truely periodic, the data-part of the frames introduce symbols randomly choosen from a uniform distribution. We split the results by modulation index:
If the modulation index $\eta$ is integer, $\eta p_{m}\in\mathbb{Z}$ and we get $e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}=1.$
Thus,
$$
\left|\rho\left[x\right]\left(\tau_{0}\right)\right|=p_{\%}\left(N-\tau_{0}+1\right).$$
If $\eta$ is not integer,
$$\left|e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right|=1$$
but the argument of the summands in ${arg}\left(e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right)$
is not constant, as it depends on the random data frame, and thus consists of a sum of complex exponentials of differing phase. The sum decreases compared to the magnitude of the sum of an integer modulation index signal. I conclude that a random walk is in order to determine the effects of the cancellation.
