Step 1: The probability of the game never ending is zero.
Let $p_n$ be the probability that Bob is eventually $n$ wins ahead of Alice. In order for this to happen, the random walk of the net number of wins for Bob needs to reach $+n$ before it reaches $-1$. By the famous Gambler's ruin problem, $$p_n=\frac{1}{n+1}$$
Now, let $B_n$ be the event that Bob is eventually ahead by $n$ wins.
$$
\begin{align}
P(\text{game goes forever})
&=P(\text{game goes forever}\mid \text{$B_n$})P(B_n)+P(\text{game goes forever}\mid B_n^c)P(B_n^c)
\\&\le 1\cdot (1/n) + P(\text{game goes forever}\mid B_n^c)\cdot 1
\end{align}
$$
I claim that $P(\text{game goes forever}\mid B_n^c)=0$. Indeed, given $B_n^c$, the random walk will never reach $+n$, so the game can only go forever if it stays in the range $\{0,1,2,\dots,n-1\}$ forever. However, during the course of any $n$ steps, there is a probability of at least $(1/2)^n$ that Alice will get $n$ wins a row, implying she wins. By dividing the infinite sequence of flips into disjoint sections of $n$, these occurrences are independent, so it will occur at least once with probability one.
Putting this altogether, the probability of the game going forever is at most $1/(n+1)$ for all $n\ge 2$, so the probability must be zero.
Step 2: The expected length of the game is infinite.
Suppose Bob wins the first game. Let $N$ denote the number of games it takes for Alice to regain the majority. We use the well known fact that $E[N]=\sum_{k=0}^\infty P(N>k)$. When $k$ is even, one way for $P(N>k)$ to occur is for both Bob and Alice to win exactly $k/2$ games in the first $k$ games, in such a way that the net number of wins for Bob stays nonnegative. Such paths are known to be enumerated by the Catalan numbers, so the probability of this occurring is
$$
\frac1{k/2+1}\binom{k}{k/2} \cdot 2^{-k}\sim \frac{C}{\sqrt k}
$$
We have shown that $E[N]=\sum_{k=0}^\infty P(N>k)\ge \sum_{k=0,\text{ $k$ even}} \frac{C}{\sqrt k}$. This last sum diverges, so $E[N]=\infty$.
For your follow-up question, if the coin has a probability $p>\frac12$ of heads, then the probability the game goes forever is $(1-p)/p$. Proven here: Hitting probability of biased random walk on the integer line