If you encrypt the messages $m_1$ and $m_2$ with the pad $p$ as
$$\begin{aligned} c_1 &= m_1 \oplus p, \\ c_2 &= m_2 \oplus p, \end{aligned}$$
where $\oplus$ denotes the binary operation of a finite group (e.g. addition on integers modulo $n$, or XOR on bitstrings, etc.) and $p$ is a random element of the group, then, indeed, an attacker who intercepts only $c_1$ and $c_2$ will not be able to recover either $m_1$ or $m_2$.
However, the attacker can recover
$$\begin{aligned} c_1 \oplus c_2^{-1} &= m_1 \oplus p \oplus (m_2 \oplus p)^{-1} \\ &= m_1 \oplus p \oplus p^{-1} \oplus m_2^{-1} \\ &= m_1 \oplus m_2^{-1}, \end{aligned}$$
where $x^{-1}$ denotes the group inverse of $x$.
(It's somewhat interesting to note that the group does not even need to be abelian for this to work: all we need is a practical way to compute inverses and apply the group operation.)
Thus, the attacker does gain information about the relationship between $m_1$ and $m_2$. In particular, if they later find out either of the messages, they will know the other one too — and, more generally, any information the attacker obtains about one of the messages will give them information about the other one.
Essentially, if the keyspace (i.e. the group from which $p$ was randomly chosen) has $n$ elements, then knowing $c_1 \oplus c_2^{-1} = m_1 \oplus m_2^{-1}$ narrows the number of possible values of $(m_1, m_2)$ from $n^2$ to $n$.
All of the above holds regardless of how the messages $m_1$ and/or $m_2$ are chosen. If both (or even just all but one) of them are completely random, and if you can guarantee that the attacker will never gain any information about the random messages beyond what they've obtained from the ciphertexts, then this knowledge will, indeed, do them no good. But, as others have pointed out, this pretty much rules out using the messages for anything. (In particular, using $m_1$ and $m_2$ as one-time pads to transmit further messages would be a very bad idea.)
