Is Poly1305 an information-theoretically secure MAC?
Can the $AES_k(n)$ portion be simply replaced with $k \oplus n$?


No, but you're close, it would be replaced with $k + n$, where $+$ is addition modulo $2^{128}$; then it becomes informational theoretic.

Here's why: Poly1305 is based on a polynomial universal hash.  This is a hash where we select a finite field $GF(p^i)$, select a private value $x \in GF(p^i)$, interpret the message being authenticated as a series of value $m_a, m_{a-1}, ..., m_1$, and compute the value:

$$m_a x^a + m_{a-1} x^{a-1} + ... + m_1 x^1 + f(k)$$

where the above additions and multiplications are done in $GF(p^i)$, and $f(k)$ is the 'unknown encrypted nonce' ($AES_k(n)$ in the Poly1305, $k$ in your case).

The idea here is if someone is able to find a second message and tag that authentications with the same nonce, then we can 'subtract' the two polynomials, and find a polynomial in $x$ that evaluates to 0.  And because $GF(p^i)$ is a field, that polynomial has no more than $a$ zeros; that is, unless $x$ is one of $a$ out of $p^i$ possible values, that polynomial won't be zero).

Now, with Poly1305, they use the values $p = 2^{130}-5$ and $i = 1$.  What this means is that the operation that adds $k$ is not bitwise xor; instead, it is modular addition.  Now, the above proof assumes that the addition is actually modulo $2^{130}-5$, not $2^{128}$; it turns out that it isn't hard to show that this increases the possible number of 0's by a relatively small factor.  I don't know of any corresponding proof if you replace the addition with xor, and hence using xor would appear to be foolhardy.