Here we shall only discuss general relativistic diffeomorphism-invariant matter theories in a curved spacetime in the classical limit $\hbar\to 0$ for simplicity. In particular, we will not discuss the SEM pseudotensor for the gravitational field, but only the stress-energy-momentum (SEM) tensor for matter (m) fields $\Phi^A$. We emphasize that our conclusions will be independent of whether the EFEs are satisfied or not.
I) On one hand, the basic Hilbert/metric SEM tensor-density$^1$ is manifestly symmetric$^2$
$$\tag{1} \sqrt{|g|}T^{\mu\nu}~\equiv~{\cal T}^{\mu\nu}~:=~-2\frac{\delta S_m}{\delta g_{\mu\nu}},$$
cf. a comment by Lubos Motl. However, note that the basic definition (1) is not applicable to e.g. fermionic matter in a curved spacetime, cf. Section II.
Diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity. Using the matter eqs. of motion (eom)
$$\tag{2} \frac{\delta S_m}{\delta \Phi^A}~\stackrel{m}{\approx}~0, $$
the corresponding Noether's 2nd identity reads$^3$
$$\tag{3} \nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0, $$
cf. e.g. Ref. 1. [Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eoms. The connection $\nabla$ is the Levi-Civita connection.] Eq. (3) serves as an important consistency check. A matter source $T^{\mu\nu}$ to the EFEs should satisfy eq. (3), cf. the (differential) Bianchi identity.
II) In the Cartan formalism, the fundamental gravitational field is not the metric tensor $g_{\mu\nu}$ but instead a vielbein $e^a{}_{\mu}$. The generalized Hilbert SEM tensor-density is defined as
$$\tag{4}{\cal T}^{\mu}{}_{\nu}~:=~{\cal T}^{\mu}{}_a~e^a{}_{\nu}, \qquad
{\cal T}^{\mu}{}_a ~:=~- \frac{\delta S_m}{\delta e^a{}_{\mu}}, $$
which is no longer manifestly symmetric, cf. e.g. Ref. 2.
Next we have two symmetries: local Lorentz symmetry and diffeomorphism invariance.
Firstly, local Lorentz symmetry leads (via Noether's 2nd theorem) to an off-shell identity. Using the matter eoms (2), the corresponding Noether's 2nd identity reads
$$\tag{5} {\cal T}^{\mu\nu}~\stackrel{m}{\approx}~{\cal T}^{\nu\mu},$$
i.e the generalized Hilbert SEM tensor-density (4) is still symmetric when the matter eoms are satisfied.
Secondly, diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity$^4$
$$\tag{6} d_{\mu}{\cal T}^{\mu}{}_{\nu}
~=~{\cal T}^{\mu}{}_a ~d_{\nu}e^a{}_{\mu}
-\frac{\delta S_m}{\delta \Phi^A}~d_{\nu}\Phi^A. $$
Not surprisingly, eqs. (5), (6), and $(\nabla_{\nu} e)^a{}_{\mu}=0$ imply eq. (3).
III) On the other hand, the canonical SEM tensor-density
$$\tag{7} \Theta^{\mu}{}_{\nu}
~:=~\frac{\partial {\cal L}_m}{\partial\Phi^A_{,\mu}}\Phi^A_{,\nu} -\delta^{\mu}_{\nu}{\cal L}_m$$
is not always symmetric, cf. e.g. this Phys.SE post. The fact that the Lagrangian density ${\cal L}_m$ has no explicit spacetime dependence leads (via Noether's 1st theorem) to an off-shell identity
$$\tag{8} d_{\mu}\Theta^{\mu}{}_{\nu}
~=~-\frac{\delta S_m}{\delta e^a{}_{\mu}}~d_{\nu}e^a{}_{\mu}
-\frac{\delta S_m}{\delta \Phi^A}~d_{\nu}\Phi^A. $$
Recall that the gravitational field, the vielbein $e^a{}_{\mu}$, is not necessarily on-shell. Remember we are doing FT in curved spacetime rather than GR. As a consequence, the first term on the right-hand side of Noether's 1st identity (8) breaks the usual interpretation of Noether's 1st theorem as leading to an on-shell conservation law. [It is comforting to see that it gets restored for a constant vielbein with $d_{\nu}e^a{}_{\mu}=0$.]
IV) Eqs. (4), (6), and (8) imply that
$$\tag{9} d_{\mu}({\cal T}^{\mu}{}_{\nu}-\Theta^{\mu}{}_{\nu})~=~0.$$
One may show that there in general exists a Belinfante improvement tensor-density
$$\tag{10} {\cal B}^{\lambda\mu}{}_{\nu}~=~-{\cal B}^{\mu\lambda}{}_{\nu},$$
such that
$$\tag{11} {\cal T}^{\mu}{}_{\nu}-\Theta^{\mu}{}_{\nu}
~=~d_{\lambda}{\cal B}^{\lambda\mu}{}_{\nu},$$
cf. e.g. my Phys.SE answer here. Note that eqs. (10) and (11) are consistent with eq. (9).
V) Eq. (11) serves as an important consistency check of the Hilbert SEM tensor-density (4) vs. the canonical SEM tensor-density (7). Eq. (11) implies that the two corresponding Noether charges, the energy-momentum $4$-covectors
$$\tag{12} P_{\nu} ~:=~ \int\! d^3x~{\cal T}^0{}_{\nu} \quad\text{and}\quad \Pi_{\nu} ~:=~ \int\! d^3x~\Theta^0{}_{\nu} $$
are equal up to spatial boundary terms
$$\tag{13} P_{\nu}-\Pi_{\nu}~=~\int\! d^3x~d_i{\cal B}^{i0}{}_{\nu}~\sim~0,$$
cf. the divergence theorem.
References:
R.M. Wald, GR; Appendix E.1.
D.Z. Freedman & A. Van Proeyen, SUGRA, 2012; p. 181.
--
$^1$ A tensor-density ${\cal T}^{\mu\nu}= e T^{\mu\nu}$ is in this context just a tensor $T^{\mu\nu}$ multiplied with the density $e=\sqrt{|g|}$.
$^2$ Conventions: In this answer, we will use $(+,-,-,-)$ Minkowski sign convention. Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.
$^3$ Note that eq. (3) is not a conservation law by itself. To get a conservation law, we need a Killing vector field, cf. e.g. my Phys.SE answer here.
$^4$ Here we have assumed that the matter fields $\Phi^A$ only carry flat or spinorial indices, cf. the setting of my Phys.SE answer here. If $\Phi^A$ also have curved indices, there will be further terms in eq. (6) proportional to the matter eoms.
Best Answer
I like to answer these kinds of questions with a simple working example. So let's write a simple scalar field theory that breaks rotation but not translation invariance: \begin{equation} \mathcal{L}= -\frac{1}{2} \left( \eta^{\mu\nu} + \lambda e^\mu e^\nu \right) \partial_\mu \phi \partial_\nu \phi \end{equation} where $\lambda$ is a constant and $e^\mu$ is some fixed unit norm ($\eta_{\mu\nu} e^\mu e^\nu=1$) spacelike 4-vector that is constant over spacetime.
Now let's do a variation with a spacetime dependent shift $\epsilon(x)$; we will then use the trick that because translations with $\epsilon={\rm const}$ are a symmetry we can always write the variation as $\delta \mathcal{L}=-\partial_\mu \epsilon_\nu \tilde{T}^{\mu\nu}$, which after an integration by parts is $\epsilon_\mu \partial_\nu \delta T^{\mu\nu}$ (where $\tilde{T}$ and $T$ potentially differ by a term whose divergence is identically zero). Then since $\partial_\mu T^{\mu\nu}=0$ on shell, we can identify $T^{\mu\nu}$ as the stress-energy tensor.
More concretely, we write $x^\mu \rightarrow x^\mu - \epsilon^\mu(x)$, so $\phi \rightarrow \phi + \epsilon^\mu \partial_\mu \phi$ and $\partial_\mu \phi \rightarrow \partial_\mu \phi + \epsilon^\nu \partial_\mu \partial_\nu \phi + \partial_\nu \phi \partial_\mu \epsilon^\nu$, leading to
\begin{eqnarray} % \delta \mathcal{L} &=& - \frac{1}{2} \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] \left[ \delta(\partial_\rho \phi) \partial_\sigma \phi + \partial_\rho \phi \delta (\partial_\sigma \phi) \right] \\ % &=& - \frac{1}{2} \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] \left[ \epsilon^\nu \partial_\rho \partial_\nu \phi \partial_\sigma \phi + \epsilon^\nu \partial_\sigma \partial_\nu \phi \partial_\rho \phi + \partial_\rho \phi \partial_\nu \phi \partial_\sigma \epsilon^\nu + \partial_\sigma \phi \partial_\nu \phi \partial_\rho \epsilon^\nu \right] \\ % &=& - \frac{1}{2} \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] \left[ \epsilon^\nu \partial_\nu \left( \partial_\rho \phi \partial_\sigma \phi \right) - \epsilon_\mu \partial_\sigma \left(\partial^\mu \phi \partial_\rho \phi \right) - \epsilon_\mu \partial_\rho \left(\partial^\mu \phi \partial_\sigma \phi \right) \right] \\ % &=& \epsilon_\mu \partial_\nu \left[ \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} (\partial \phi)^2 + \lambda \left( e^\mu \partial^\nu \phi e^\rho \partial_\rho \phi - \frac{1}{2} e^\mu e^\nu (\partial \phi)^2\right) \right]\\ % &=& \epsilon_\mu \partial_\nu \left[T_{\rm K.G.}^{\mu\nu} + \lambda T_{\rm S.B.}^{\mu\nu} \right] \end{eqnarray} where $T_{\rm K.G.}^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu}(\partial \phi)^2$ is the stress tensor for a massless Klein-Gordon field, and the term $T^{\mu\nu}_{\rm S.B.}$ is the stress-energy tensor due to the breaking of rotational symmetry (S.B.=symmetry breaking).
Indeed, the expression for $T^{\mu\nu}_{\rm S.B.}$ \begin{equation} T^{\mu\nu}_{\rm S.B.} = e^\mu \partial^\nu \phi e^\rho \partial_\rho \phi - \frac{1}{2} e^\mu e^\nu (\partial \phi)^2 \end{equation}
has a non-vanishing anti-symmetric part $\sim (e \cdot \partial)\phi e^{[\mu}\partial^{\nu]} \phi $. One way to explain in words what is happening is that $\epsilon_\mu$ is never directly contracted with $e^\rho$ or $e^\sigma$, and unlike the metric tensor there is no way to "generate more" $e^\rho$ or $e^\sigma$ by raising and lowering indices. Another way is to explain the difference between the rotationally invariant and symmetry breaking terms is to point out that if we replace $e^\mu e^\rho$ with $\eta^{\mu\rho}$ in the first term of $T^{\mu\nu}_{\rm S.B.}$, the first term becomes $\partial^\mu\phi \partial^\nu \phi$, which is symmetric; the "symmetry breaking" term $\sim e^\mu e^\rho$ picks out a special direction on which the final answer can depend, which allows for an anti-symmetric part. To be honest, this isn't the result I was expecting when I started working this out (although I don't know why since there is no way angular momentum can be conserved and I agree with the logic in your question), but as far as I can tell it is correct.
It's always possible to define a symmetric stress-energy tensor by defining it as a variation of the action with respect to the metric, but I take it from the question that this is not what you are interested in.
Another question is whether it is possible to add piece that is identically divergence free to $T^{\mu\nu}_{\rm S.B.}$ to cancel the anti-symmetric part... I suspect (but am not 100% sure) the answer is yes, since I would think you should be able to derive the Einstein-Hilbert version of the stress-energy tensor in this way. (This isn't always possible but I suspect it probably is in this example)