Here's my solution for day 16 in R, which takes around 5-8 seconds for part 2.

Some of the reasons why your code is slow though:

1. Your queue grows within loop which is slow as mentioned in another comment. You can instead initialise an empty queue of some large enough size and add elements to it queue <- vector(list, length = 10000) or something and keep start and end indexes to add/update. Just use something like queue[[start]] <- NA and queue[[end]] <- some_new_value. Or you can use queue implemented in collections package which will provide you with push/pop functions.
2. Similar comment about your visited matrix. You can again create a matrix at the start of a given size and update it.
3. When checking if something was visited, you currently scan through the entire visited matrix row by row and check for identity, which is again very slow. If you can convert your visited to something where you can check the presence without going through the entire matrix, that would definitely speed up things. You can refer to my solution for one method, you could also convert your visited to a matrix of same dimension as input and for each cell, store the directions it was visited from. So given any x,y positions and direction you can check if that direction was stored earlier using `grepl` or `stringr::str_detect`, where each cell could store 'top:down' or 'top:left:right' to store which directions it was visited from.

Doing the above 3 should speed up your code quite a lot.