Is it possible to permutate a periodic sequence in such a way that any infinite substring will always be non-periodic?

I am a bit confused by your term substring, but I will give you several different constructions that will answer your question.

If you are looking for a permutation such that for any infinite subsequence (that is you are allowed to skip elements of the sequence), then this is impossible as since if the digit $x$ appears in the sequence, then it will appear infinitely often in the sequence, so after the permutation, you can just take the subsequence that just contains the digits $x$.

If $(x_n)$ is your periodic sequence and $(y_n)$ is the permutation, if you are curious if there is a permutation of $(y_n)$ such that for any $k$ the sequence $(y_n)_{n>k}$ (where you cut off the first $k$ terms) is not periodic. Then I claim that this can be done if and only if the initial sequence you started with is not the constant sequence. Consider the following construction. Say that $(x_n)$ has some period $k$ so we have that the sequence is $x_1,x_2,...,x_k,x_1,x_2,...,x_k,x_1,...$. Since it is not the constant sequence there exists different elements of the sequence say $x_i\neq x_j$, let us denote the elements appearing $z_1,...,z_l$ where $1<l\leq k$, then since the sequence $(x_n)$ is periodic we have that each $z_i$ appears infinitely often in the sequence, so let us form the sequence $$z_1,z_2,...,z_l,z_1,z_1,z_2,z_2,...,z_l,z_l,z_1,z_1,z_1,...$$ If you consider this sequence there is no infinite subsequence that will be periodic as you go out farther there will be arbitrarily long substring of the form $z_1,z_1,...,z_1$.