Can we make sense of the logarithm of prime in some algebraic extension of $\mathbb{Q}_q$, where either $q \neq p$ or $p = q$ and both prime numbers?
Some reflections: A naive starting point is perhaps something like $$ \log( 1+x) = -\sum\limits_{n=1}^{\infty} \frac{(-x)^n}{n},$$ which converges in the reals in a neighborhood of $|x|_\infty \leq 1$, but $|p+1|_q$ might be small, at least for $q=p$ it is actually $|p+1|_p=1$, hence it lies at the boundary of the circle of convergence, where I assume that the radius remains by $1$ in the $q$ adic world. Are there analogues of Abel's theorem for the convergence at the boundary?
I think that you found the main obstacle for this to work. If you look at the exponential function $$ E(x)=1+\sum_{n=1}^\infty\frac{x^n}{n!} $$ you see that it cannot converge unless $|x|_q<1$, because in this case the denominator makes things worse (in sharp contrast to the archimedean case). Actually we need a little bit more than this, because the $q$-adic value of the factorial tends towards zero. A more careful analysis starting from the fact that $|n!|_q=q^{-t}$, where $$ t=\left[\frac{n}{q}\right]+\left[\frac{n}{q^2}\right]+\left[\frac{n}{q^3}\right]+\cdots= \sum_{k=1}^{\lceil\log_q n\rceil}\left[\frac{n}{q^k}\right] $$ reveals that we need $|x|_q<q^{-1/(q-1)}$ for $E(x)$ to converge.
So if $x\in\mathbf{Q}_q$, then we must have $x\in q\mathbf{Z}_q$ for $E(x)$to converge, and then $E(x)\equiv 1\pmod{q\mathbf{Z}_q}$. Unless I made a mistake, seeking an $x$ from an extension field is not going to change this.
So in order for the logarithm to make sense, the congruence $p\equiv 1\pmod{q}$ that you observed is necessary and sufficient. See also http://en.wikipedia.org/wiki/P-adic_exponential_function for more discussion, links, and workarounds.