Does $\pi$ span an infinite grid?
I suspect your friend is thinking as follows: the digits of $\pi$ are "basically random," so the walk generated by $\pi$ should go "all over the place." There are two pieces to this to consider:
- How random is $\pi$?
and
- How do random walks behave?
Let me begin with the second question, because it's easier to answer. We forget about the digits of $\pi$, and instead imagine we have some fair die that we're rolling, with an appropriate number of sides, and using this to generate a random walk. What happens?
Let's ignore base $8$ for the moment; the bases $4$ and $2$ are easier to visualize. In base $2$, we move left/right in a line; in base $4$, we move up/down/left/right in the plane. In either case, we're moving in a particularly nice lattice. Similarly, the base $6$ version is a cubical lattice (up/down/left/right/forward/back), and so on. In general, the base $2n$ version is a walk in the $n$-dimensional cubical lattice.
So fix $n$, and roll a $2n$-sided die repeatedly. What's the probability that we "fill the space?"
Surprisingly, the dimension matters! Specifically, if $n=1$ or $n=2$ the answer is $1$ (we almost certainly fill the space), whereas if $n\ge 3$ the answer is $0$ (we almost certainly don't fill the space). See Polya's recurrence theorem.
So if we believe $\pi$ "behaves randomly" (whatever that means, for the moment), then we should believe that the base-4 version of $\pi$ fills the plane, and the base-2 version fills the line; but the base-6 version probably doesn't fill 3d-space, and so on.
What about the base-$8$ version you have, which doesn't really fit into this cubical pattern? Well, it will still fill space with probability $1$, but that takes a bit of work.
OK, so what about the second question - how random is $\pi$?
Say that a real number $r$ is simply normal to base $b$ if, in the base-$b$ expansion of $r$, each of the possible $b$-many digits appears ${1\over b}$th of the time. (That is, $$\lim_{n\rightarrow\infty}{\mbox{number of occurrences of $d$ in the first $n$ base-$b$ digits of $r$}\over n}= {1\over b},$$ for each digit $d$.) For example, Champernowne's constant is simply normal to base $10^k$ for every $k$. A number $r$ is absolutely normal if it is simply normal to every base; this implies that every sequence of $k$ digits in base $b$ appears the "right" proportion of the time in the base-$b$ expansion of $r$ (consider base $b^k$).
It is strongly believed that $\pi$ (and many other interesting real numbers) are absolutely normal, but this remains wildly open. Currently, we do not even know a single "natural" example of a real number which is absolutely normal! (We do know explicit examples - see e.g. this 2002 paper of Becher - but they are very contrived.) However, we also know that "most" (measure-$1$) real numbers are absolutely normal! Basically, if you determine the digits of a real number by flipping an appropriate coin, with probability $1$ the result is absolutely normal.
To the best of my knowledge, it is not known whether $\pi$ is simply normal to any base, or even if every digit occurs infinitely often in $\pi$ for in any specific base $>2$. As the comments above have stated, absolute normality is almost certainly overkill for showing that the 2d-walk fills the plane with probability $1$ (see Robert Israel's answer below), but I don't think we have proved any sufficient amount of "randomness" about $\pi$.
Equiprobability in base $8$ is not necessary, as you could tour the plane with e.g. digits $1,4,6$.
And normality is not sufficient: consider a sequence that is random except that when the walker would otherwise hit $[1,0]$ you change the digit. Although this will require changing infinitely many digits (because a random walker in the plane will hit $[1,0]$ infinitely many times), the fraction of digits to be changed goes to $0$ (asymptotically, the probability of the random walker being at $[1,0]$ at time $t$ decreases like $1/t$), so that these changes do not affect normality in any base.