1) There is an escape route that touches no cell more than once for every position in a finite (n-cell) maze, with each escape route being n-cells long or less. The way to see this is just to grow multiple strands of a string from the exit inward toward every cell, bifurcating at each maze branch, so as to reach all cells. The strands are the paths out. Thus, for every starting position there is a "way out" consisting entirely of "make-able" moves which traverses n or fewer cells. In the same way, there is a sequence of n or less moves from any one cell of the maze to any other. For a maze of n cells, the set of all possible n or less-length sequences may be generated. The total number of moves in this sequence of sequences is the sum of the number of sequences of each length  (4^1 + 4^2 + ... + 4^n ). Call this sum f(n).  

2) There is no possibility for G to build-in to G's sequence "retracing instructions" for each failed path, since blockages are irreproducible when backtracking and so backtracking it impossible. This means that G cannot build in a "reverse this attempt" sequence following every sequence of moves. So, each sequence brings G to an unknowable position, likely one different than the previous position. 

3) A note on infinities: Countable infinities multiplied together still make countable infinities. For a well known example, consider a set containing the positive integers, and consider a set of these sets indexed by the positive integers. How many integers are there in total? This total is equivalent to the number of integer grid points in Quadrant I of the coordinate plane, which may be counted on a zig-zag line starting at the origin. Thus, there are still a countable infinity of elements in the product set.  

4) A note on probabilities: If there is a non-zero probability of an event occurring in a random process, and an infinite number of realizations of the process are considered, then eventually (in a finite time) that event will occur. As in QM: that which is not forbidden is compulsory. 

For a beginning idea of where this is going, image a maze of n cells.

G makes a list of all f(n) sequences and concatenates them in an infinite number of random orderings so that each of the f(n) members appear an infinite number of times in what we'll call the random Grand Sequence. Note that the length of the GS is the product of two countable infinities.  This GS will, by chance, eventually match the phase of the wanderer. In other words, G, moving through the maze will eventually be in the right cell when the right route out starts. The reason for this is that while every exit sequence is in the GS, so is every path from every cell to every other cell. By chance, G will be shuttled to every cell and also by chance, at some point in a finite amount of time, find that the appropriate exit sequence will start when G is in one of the cells. Think of it like this: G is constantly being showered with sequences and is also constantly moving through the maze. Eventually the right sequence for the exit will hit G when G is in the right cell for it.  

The final wrinkle in this solution is that the actual size of n is unknown. So G's GS must be built by concatenating the 4^n sequences from ever larger and larger n while also randomly concatenating-in sequences from 4^m where 1<m<n. By ordering them randomly and endlessly, even with n growing, the right sequence matching G's position will be mode.  (This is the multiplication of a third countable infinity to make the GS.)

Finally, we ask: could one completely random infinite sequence of U, D, R, and Ls contain all the aspects required in the GS? I.e, skip making the GS entirely? Again, the answer is yes. An infinite base 4 number of random digits is necessarily a "normal number" that contains any desired pattern, including all f(n) sequences in all random orders out to any desired finite length. This very long random number, easy to produce, contains all f(n) arranged randomly an infinity of times for every n, no matter how large is n. (That which is not forbidden is compulsory).  This random number alone is all G needs to get out. So, this then is the simplest way out for G.