For a regular graph, each walk of a given length has the same probability, so let's just consider the number of walks.
A walk starting and ending at a given vertex is comprised of zero or more pieces that consist of non-trivial walks that return to the start only on their last step. So if $w(x)$ is the ordinary generating function of all walks that end at the starting vertex, and $r(x)$ is the ordinary generating function of the non-trivial walks that first return to the start on the last step, then
$$ w(x) = 1 + r(x) + r(x)^2 + \cdots = \frac{1}{1-r(x)},$$
or equivalently
$$ r(x) = \frac{w(x)-1}{w(x)}.$$
If you know the coefficients of $w(x)$, this lets you get the coefficients of $r(x)$.
Since these generating functions are rational functions (see Didier's answer for a proof), the asymptotics of the probability you want are determined by the smallest (complex) solution of $w(x)=0$.
First, let me elaborate on my comment. If $X$ occurs with probability $\rho$, then the expected waiting time before you first see a streak of length $\ell$ $X$s in a row in independent trials is exponential in $\ell$, $\frac{\rho^{-\ell}-1}{1-\rho}$. If you get $2n-1$ right steps in a row, then you must have hit the barrier, which gives an exponential upper bound for the expected first hitting time, $\frac{p^{1-2n}-1}{q+r}$. On the other hand, to reach a barrier, you need a streak of at least $n$ non-deaths. Again, the waiting time before you first see such a streak is exponential for $r\gt 0$, $\frac{(p+q)^{-n}-1}{r}$. If $q=0$ or $p=0$, the lower bound is sharp.
Imagine we release a particle at $0$ once, and watch it until it reaches a barrier or dies. Let the average number of steps it takes be $s$, and let the probability that it dies be $d$. Then the expected number of steps before some particle hits a barrier is $s(1+d+d^2+...) = \frac{s}{1-d}$. Both $s$ and $d$ can be calculated analytically, but I think it's easier to let Mathematica do it. Here are Mathematica commands which find them:
ProbReachesBarrier[k_] := Evaluate[a[k] /.
RSolve[{a[k] == p a[k + 1] + q a[k - 1], a[n] == 1, a[-n] == 1},
a[k], k][[1]] ]
ProbDies = Simplify[1 - ProbReachesBarrier[0]]
Let $\alpha = \frac{1+\sqrt{1-4pq}}{p}, \beta = \frac{1-\sqrt{1-4pq}}{p}$.
$$d = - \frac{(2^n-\alpha^n)(2^n-\beta^n)}{2^n(\alpha^n + \beta^n)}$$
ESteps[k_] := Evaluate[b[k] /.
RSolve[{b[k] == 1 + p b[k + 1] + q b[k - 1], b[n] == 0, b[-n] == 0},
b[k], k][[1]] ]
s = Simplify[ESteps[0]]
$$ s = - \frac{(2^n - \alpha^n)(2^n-\beta^n)}{r 2^n (\alpha^n + \beta^n)} = \frac{d}{r}$$
$$\frac{s}{1-d} = - \frac{(2^n-\alpha^n)(2^n-\beta^n)}{r (4^n + \alpha^n \beta^n)}.$$
Best Answer
The explicit formula is: $P[N_m=n]=(m/n)P[S_n=m]$, where $P[N_m=n]$ is the probability the position $m$ is hit after exactly $n$ steps, $S_n = X_1+X_2+\dots X_n$ and $P[S_n=m]$ is the probablity after $n$ steps the path to be at the position $m$. This last is well-known and is given by
$$P[S_n=m]=\frac{n!}{[(n+m)/2]![(n-m)/2]!}p^{(n+m)/2}q^{(n-m)/2}.$$
This is true when $n$ and $m$ have the same parity, else the probability is zero. Additionally $m$ is positive and $n$ is greater than or equal to $m$. A similar expression can be found for negative $m$.