There are several misconceptions here, most of them originate from exposure to only the *pure* state formalism of quantum mechanics, so let's address them one by one:

All quantum operations must be unitary to allow reversibility, but
what about measurement?

This is false. In general, the states of a quantum system are not just vectors in a Hilbert space $\mathcal{H}$ but density matrices $-$ unit-trace, positive semidefinite operators acting on the Hilbert space $\mathcal{H}$ i.e., $\rho: \mathcal{H} \rightarrow \mathcal{H}$, $Tr(\rho) = 1$, and $\rho \geq 0$ (Note that the pure state vectors are not vectors in the Hilbert space but **rays in a complex projective space**; for a qubit this amounts to the Hilbert space being $\mathbb{C}P^1$ and not $\mathbb{C}^2$). Density matrices are used to describe a statistical ensemble of quantum states.

The density matrix is called **pure** if $\rho^2 = \rho$ and **mixed** if $\rho^2 < \rho$. Once we are dealing with a pure state density matrix (that is, there's no statistical uncertainty involved), since $\rho^2 = \rho$, the density matrix is actually a projection operator and one can find a $|\psi\rangle \in \mathcal{H}$ such that $\rho = |\psi\rangle \langle\psi|$.

The most general quantum operation is a CP-map (completely positive map), i.e., $\Phi: L(\mathcal{H}) \rightarrow L(\mathcal{H})$ such that $$\Phi(\rho) = \sum_i K_i \rho K_i^\dagger; \sum_i K_i^\dagger K_i \leq \mathbb{I}$$ (if $\sum_i K_i^\dagger K_i = \mathbb{I}$ then these are called CPTP (completely positive and **trace-preserving**) map or a quantum channel) where the $\{K_i\}$ are called Kraus operators.

Now, coming to the OP's claim that all quantum operations are unitary to allow reversibility -- this is just not true. The unitarity of time evolution operator ($e^{-iHt/\hbar}$) in quantum mechanics (for closed system quantum evolution) is simply a consequence of the Schrödinger equation.

However, when we consider density matrices, the most general evolution is a CP-map (or CPTP for a closed system to preserve the trace and hence the probability).

Are there any situations where non-unitary gates might be allowed?

Yes. An important example that comes to mind is open quantum systems where Kraus operators (which are not unitary) are the "gates" with which the system evolves.

Note that if there is only a single Kraus operator then, $\sum_i K_i^\dagger K_i = \mathbb{I}$. But there's only one $i$, therefore, we have, $K^\dagger K = \mathbb{I}$ or, $K$ is unitary. So the system evolves as $\rho \rightarrow U \rho U^\dagger$ (which is the standard evolution that you may have seen before). However, in general, there are several Kraus operators and therefore the evolution is non-unitary.

Coming to the final point:

Measurement can be represented as a matrix, and that matrix is applied to qubits, so that seems equivalent to the operation of a quantum gate. That's definitively not reversible.

In standard quantum mechanics (with wavefunctions etc.), the system's evolution is composed of two parts $-$ a *smooth* unitary evolution under the system's Hamiltonian and then a *sudden* quantum jump when a measurement is made $-$ also known as wavefunction collapse. Wavefunction collapses are described as some projection operator say $|\phi\rangle \langle\phi|$ acting on the quantum state $|\psi\rangle$ and the $|\langle\phi|\psi\rangle|^2$ gives us the probability of finding the system in the state $|\phi\rangle$ after the measurement. Since the measurement operator is after all a projector (or as the OP suggests, a matrix), shouldn't it be linear and physically similar to the unitary evolution (also happening via a matrix). This is an interesting question and in my opinion, difficult to answer physically. However, I can shed some light on this mathematically.

If we are working in the modern formalism, then measurements are given by POVM elements; Hermitian positive semidefinite operators, $\{M_{i}\}$ on a Hilbert space $\mathcal{H}$ that sum to the identity operator (on the Hilbert space) $\sum _{{i=1}}^{n}M_{i}=\mathbb{I}$. Therefore, a measurement takes the form $$ \rho \rightarrow \frac{E_i \rho E_i^\dagger}{\text{Tr}(E_i \rho E_i^\dagger)}, \text{ where } M_i = E_i^\dagger E_i.$$

The $\text{Tr}(E_i \rho E_i^\dagger) =: p_i$ is the probability of the measurement outcome being $M_i$ and is used to *renormalize* the state to unit trace. Note that the numerator, $\rho \rightarrow E_i \rho E_i^\dagger$ is a linear operation, but the probabilistic dependence on $p_i$ is what brings in the *non-linearity* or *irreversibility*.

Edit 1: You might also be interested Stinespring dilation theorem which gives you an isomorphism between a CPTP map and a unitary operation on a larger Hilbert space followed by partial tracing the (tensored) Hilbert space (see 1, 2).

2I've downvoted this because it refers to a concept which is arguably more advanced than the perspective from which the original question is being asked. – Alexander Soare – 2020-05-21T11:55:44.523

5+1 Everyone interested in quantum mechanics (not just quantum information) should know about quantum operations e.g. from Nielsen and Chuang.

I think it is worth mentioning (since the Wikipedia page on Stinespring dilation is too technical) that every finite-dimensional quantum operation is mathematically equivalent to some unitary operation in a larger Hilbert space followed by a restriction to the subsystem (by the partial trace). – Ninnat Dangniam – 2018-03-20T05:31:18.690