Heuristically, one would expect the answer to be yes. There's an existing partially explicit version of this mentioned in Richard Guy's "[Unsolved Problem in Number Theory](https://doi.org/10.1007/978-0-387-26677-0)" entry F24, which is that for $n> 86$, $2^n$ always contains a zero in its base 10 expansion. There are some related questions also in that entry and some other entries in the book as well. For example, there's a conjecture that every sufficiently large power of 2 contains 0s, 1s and 2s in its base 3 expansion. The reasonable generalization (which I have not seen stated explicitly but seems to be implicit in all of these), is that if $a$ and $b$ are relatively prime integers both greater than 1, then for all sufficiently large $n$, $a^n$ has in its base $b$ expansion all of $0, 1, \dotsc b-1$. It is also plausible that the sufficiently large is a not very fast growing function, something like being true for all $n \geq a^2b^2$. Edit: See Emil's comment [below](https://mathoverflow.net/questions/423275/does-the-base-10-representation-of-2n-contain-all-10-digits-for-all-sufficien#comment1087831_423298): as written this doesn't include the case $2$ and $10$; one wants that there's a prime $p$ such that $p$ divides exactly one of $a$ and $b$. That includes the relatively prime case and also cases like $2$ and $10$. 

However, the set of $a$ and $b$ where we can prove anything like this is small and mostly relegated to when $b=2$. In that case, some of these results are implicitly very old, dating back to actually the middle ages. For example, any power of $3$ greater than $3$ must contain both $1$ and $0$ in its base $2$ expansion. This follows, since if not, $3^n$ would have to be of the form $2^k-1$, and Gersonide's theorem that $8$ and $9$ are the largest power of 2 next to a power of $3$ then applies. For larger bases, Mihailescu's proof of Catalan's conjecture shows that the same is essentially true if one has $b=2$ and $a$ is any number greater than $2$.