We shall construct a counterexample for an alphabet of $2$ letters (say, $a$ and $b$) and sufficiently large $n$ (one can compute the explicit value from various inequality constraints we will impose but it is surely far from optimal; in fact I suspect that $n=5$ is enough, though $n=2$ is, probably not).
Let $Q$ be a large odd multiple of $3$ to be chosen later. We shall construct by induction some non-empty sets $S_k$ of strings of length $Q^k$ with the property that for every $k\ge 1$,
(a) Each $s\in S_{k+1}$ is a concatenation of $Q$ strings from $S_{k}$;
(b) If $s\in S_{k+1}$, there is no $T\in\{1,\dots,Q^{k}-1\}$ such that some finite arithmetic progression of integers $t+ qT:\ q=0,\dots,n$ with difference $T$ hits every of the substrings in the property (a) and has the same number of $a$'s in $(t+qT,t+(q+1)T]$ for every $q=0,\dots, n-1$;
(c) If we view $S_k$ as a probability space (with all elements having the same probability $\frac 1{|S_k|}$) and the number $\xi=\xi(s)$ of $a$'s in the string $s\in S_k$ as a random variable, then for every integer $m$, one has $\mathcal P(\xi=m)\le 2^{-k}$.
The main claim is that if $s\in S_k$, then it cannot contain the requested pattern for $n\ge 2Q^2$. Indeed, if it does, then let $t$ be the position of the last element of $\beta$ and let $T$ be the length of $\gamma$ (in the OP notation). Choose an integer $k$ so that $Q^{k-1}\le T<Q^k$. Clearly, $k\ge 1$ and the arithmetic progression $t+qT$ cannot jump over any $Q$-adic block of length $Q^k$. Moreover, since $T\ge Q^{k-1}$, no matter where we started, in at most $Q^2$ steps we'll arrive to the beginning of some $Q$-adic block of length $Q^{k+1}$ and then in $Q^2$ more steps we'll go over all of it, thus creating an arithmetic progression that was ruled out by property (b) (observe that by property (a), every $Q$-adic block of $s$ of length $Q^{k+1}$ is in $S_{k+1}$).
Note that we haven't used property (c) in the main claim at all and it is, indeed, not needed for the conclusion. However, it is crucial for making the induction step work.
For $k=1$, we shall just take $S_1$ to be the sat of all strings of length $Q$. (a) and (b) are vacuous here and (c) immediately follows from the fact that $\mathcal P(\xi=m)=\mathcal P(\xi=Q-m)$ and $m\ne Q-m$ since $Q$ is odd (you can say much more, of course, but I try to keep it low-tech whenever possible).
Suppose that we have already built $S_k$. Consider first the set $\widetilde S_{k+1}$ of all concatenations of $Q$ strings in $S_k$. We need the following technical
Lemma. There exists $Q_0$ such that for every $Q\ge Q_0$ and every probability distribution $p$ on integers with $\|p\|_\infty\le\frac 12$, one has $\|p*p*\dots *p\|_\infty\le \frac 14 \|p\|_\infty$ where $Q$ is the number of $p$'s in the convolution.
The lemma is well-known and not terribly hard but the proof I have is a bit more high-tech than I would like it to be, so I'll wait a few days before posting it in the hope that someone will come up with a completely elementary proof.
Assuming the lemma, we conclude that the probability that the number of $a$'s in a random string from $\widetilde S_{k+1}$ is any given number does not exceed $\frac 14 2^{-k}$. We now just want to remove the bad strings $s$ from $\widetilde S_{k+1}$ to ensure (b). Note that we have $Q^k$ choices of the beginning of the bad progression (in the first $Q$-adic block of $s$ of length $Q^k$, less than $Q_k$ choices of $T$ and, for each $T$, only $T+1\le Q^k$ choices of the number of $a$'s in each interval $(t+qT,t+(q+1)T]$, thus giving us $\le Q^{3k}$ possible bad situations. Now in each particular bad situation, we can choose the intervals $(a_1, b_1],\dots (a_{Q/3},b_{Q/3}]$, with $a_j$ and $b_j$ being some elements of our arithmetic progression in $3j-2$-nd and $3j$-th $Q$-adic block of length $Q^k$ of our string $s\in \widetilde S_{k+1}$, and that we know the exact number of $a$'s in each interval $(a_j,b_j]$. However, since we can insert a arbitrary string from $S_k$ as the $3j-1$-st $Q$-adic block of $a$, the probability that this number is just right in one particular $(a_j,b_j]$
is at most $2^{-k}$. Also, since the intervals $(a_j,b_j]$ are controlled by disjoined triples of $Q$-adic blocks of length $Q^{k}$, these events are independent, so the probability that a fixed bad situation happens is at most $2^{-k\frac Q3}$. Thus the probability that some
bad situation happens is at most
$$
2^{-k\frac Q3}Q^{3k}=\left[2^{-\frac Q3}Q^{3}\right]^k\le\frac 12
$$
if $Q$ is so large that $2^{-\frac Q3}Q^{3}\le\frac 12$.
Thus we need to throw out at most one half of the full probability space $\widetilde S_{k+1}$, which would raise the probability of any event at most twice. Since having any fixed number of $a$'s in $\widetilde S_{k+1}$ had probability $\le \frac 14 2^{-k}$, we conclude that in $S_{k+1}$ it may be only twice greater, i.e., $\le 2^{-(k+1)}$, establishing (c).
That's it (modulo the technical lemma, which, as I said, I'll post later).
