This corresponds to calculating the expectation of the first-passage time of an asymmetric one-dimensional random walk. I believe that even the asymmetric one-dimensional random walk satisfies the (strong) Markov property, so therefore we can assume that the random walk "starts over" at every step. (See p. 4)
I.e. if $T(1)$ is the random variable indicating the first time we have one more head than tail, then $T(2)$ is the sum of two independent copies of $T(1)$. So by linearity of expectation: $$\mathbb{E}[T(2)]=\mathbb{E}[T(1)]+\mathbb{E}[T(1)]=2\mathbb{E}[T(1)]. $$
Now, $\mathbb{E}[T(1)]$ is a lot easier to calculate. According to this document (p.3 & p.5), $$\mathbb{E}[T(1)] = \Phi'(1)\,, \quad \text{where} \quad \Phi(t) = \frac{1 - \sqrt{1-4p(1-p)t^2}}{2(1-p)t} $$ i.e. $$\mathbb{E}[T(1)] = \frac{1 }{2p-1} \quad (p > \frac{1}{2})\,.$$ (In the document, $N= T(1)$.)