The converse does not hold. Take $X = [0,\infty)$ with Lebesgue measure. Let $a_n$ be a sequence of positive numbers such that $a_n \to 0$ but $\sum_n a_n = \infty$. (For instance, $a_n = 1/n$ would work.) Let $b_n = \sum_{i=0}^n a_k$. Then let $f_n = 1_{[b_{n-1}, b_{n})}$. Then $f_n \to 0$ a.e. and $\int f_n dm = a_n \to 0$ also. But if $f$ is a dominating function, we must have $f \ge \sup_n f_n = 1$ which is not integrable.
For a necessary and sufficient condition for convergence, look up the Vitali convergence theorem.
Let me write out my comment so that this question gets an answer. The dominated convergence theorem says the following.
If a sequence of measurable functions $\displaystyle \{f_n(x)\}_{n=1}^{\infty}$, on the measure space $(\Omega, \mathcal{F}, \mu)$, converge point-wise to a function $f(x)$ and every $f_n(x)$ is dominated by some integrable function $g(x)$ i.e. $\lvert f_n(x) \rvert \leq g(x)$, $\forall n$, $\forall x \in \Omega$ and $\displaystyle \int_{\Omega} g d\mu < \infty $, then $\displaystyle \int_{\Omega} f d \mu$ exists and $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n d \mu = \int_{\Omega} f d \mu$$
Note that it is important that the dominating function $g$ is integrable as Henry T. Horton points out in the comments. Also, the condition $\lvert f_n \rvert \leq g$ everywhere, can be relaxed to $\lvert f_n \rvert \leq g$ $\mu$-almost everywhere.
Also, we need $s > -1$. Otherwise the integrals given in the question diverge.
Now lets apply this to the problem at hand. We want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx$.
Define $$f_n(x) = \begin{cases} \left( 1 - \frac{x}n\right)^n x^s & \text{ if }x \in [0, n]\\ 0 & \text{ otherwise}\end{cases}$$
Hence, $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx$.
Now note that $f_n(x)$ converges point-wise to $e^{-x} x^s$ on $[0, \infty)$. This is so since $$\displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{x}{n} \right)^n = \exp(-x).$$
More importantly, $f_n(x)$ is dominated by $e^{-x} x^s$ on $[0,\infty)$ i.e. $\lvert f_n(x) \rvert \leq e^{-x} x^s$. This follows from the fact that $$1 - t \leq \exp(-t)$$ whenever $0 \leq t \leq 1$. Hence, we get that $$\left(1 - \frac{x}{n} \right) \leq \exp \left(-\frac{x}{n} \right)$$ which in-turn gives us $$\left(1 - \frac{x}{n} \right)^n \leq \exp \left(-x \right).$$
Hence, $\displaystyle f_n(x) \leq \exp(-x) x^s $. Also, $\displaystyle \int_0^{\infty} \exp(-x) x^s dx < \infty$ for all $s > -1$. Hence, in this case, the dominating function is the same as the limit function.
Putting all these things together, we get the desired result.
\begin{align}
\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx & = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx & \text{(From the definition of $f_n(x)$)}\\
& = \int_0^{\infty} \lim_{n \rightarrow \infty} f_n(x) dx & \text{(Since $f_n(x)$ is dominated by $f(x)$)}\\
& = \int_0^{\infty} f(x) dx & \text{(Since $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = f(x)$)}\\
& = \int_0^{\infty} \exp(-x) x^s dx
\end{align}
Best Answer
Given two positive functions $f$ and $g$ such that $f$ dominates $g$, let $$\nu(A) = \int_A g d\mu, \qquad \eta(A) = \int_A f d\mu.$$ We note that $\nu(A) \leq \eta(A)$ for all measurable sets $A$. In particular, $\nu(A)=0$ for all $\eta$-null sets $A$, so that $\nu \ll \eta$. In this sense,
Domination of functions implies absolute continuity of the induced measures.
Of course, the converse is wildly false. For example, the measures induced by $f=2$ and $g=1$ are absolutely continuous with respect to each other, whereas $f$ dominates $g$ (and not conversely).
I suppose you could use the Radon-Nikodym theorem to further analyze this situation.