Is there a basis such that if we measure the bell state $\dfrac{|00\rangle+|11\rangle}{\sqrt2}$ the results might not be correlated at all (or at least not maximally)?
For example $\dfrac{|00\rangle+|11\rangle}{\sqrt2}$ = $\dfrac{|++\rangle+|--\rangle}{\sqrt2}$ = $\dfrac{|LL\rangle-|RR\rangle}{\sqrt2}$. The results are perfectly correlated no matter in which of these bases we measure. I used this bell state as an example but I'm asking about in general with quantum states.
The property you seem to be referring to is the invariance of maximally entangled (bipartite) states upon unitary transformations of the form $U\otimes \bar U$ for any $U$. In fact, this property is characteristic of maximally entangled and maximally mixed states only. Mind the complex conjugate here: it's the reason you don't have $|00\rangle+|11\rangle=|LL\rangle+|RR\rangle$ but rather $|00\rangle+|11\rangle=|LR\rangle+|RL\rangle$.
About correlations, yes, kinda. For any choice of projective measurement basis on the first space, there's a choice on the second with respect to which you have full correlation. To see this, observe the following general property of a maximally entangled state $\sqrt2|\Psi\rangle=|00\rangle+|11\rangle$ (though the property works in higher dimensions as well): $$\langle u,v|\Psi\rangle = \bar u_0\bar v_0 + \bar u_1\bar v_1 = \langle u|\bar v\rangle,$$ holding for any pair of vectors $|u\rangle, |v\rangle$. It follows that if the first party measures in $\{|u\rangle,|u_\perp\rangle\}$ and the second one in $\{|\bar u\rangle,|\bar u_\perp\rangle\}$, then there's full correlation. Again, the conjugate is important, otherwise the statement is false eg when they both measure in the $Y$ eigenbasis, as $$\langle L,R|\Psi\rangle = \langle L|L\rangle = 1\neq0.$$
Of course, whether this makes entanglement "basis independent" depends on what precisely you mean by that sentence. Entanglement as a property isn't defined with respect to any basis, so in that sense yes, of course it's basis-independent. But that doesn't seem to be what you're asking about.
See also this related answer of mine on qc.SE.
Entanglement is generally a basis-independent concept. Suppose I have two quantum systems $A$ and $B$, with the joint system being in the pure state $\vert \psi\rangle$, for simplicity. One way of quantifying the entanglement between $A$ and $B$ is through the purity of the reduced density matrix of the $A$ subsystem $\rho_A = \mathrm{Tr}_B(\vert\psi\rangle\langle\psi\vert)$: $$S(\vert \psi\rangle) = \mathrm{Tr}(\rho_A^2),$$ where $\mathrm{Tr}_B$ denotes the partial trace over subsystem $B$. (You can also do this on the other subsystem if you want, the two turn out to be equivalent.) Basically, the smaller $S(\vert \psi\rangle)$ is, the "more entangled" $A$ is to $B$. If $S(\vert \psi\rangle)$ is minimized, the state $\vert \psi\rangle$ is maximally entangled, and conversely, if $S(\vert \psi\rangle) = 1$ (its maximum value), $\vert \psi\rangle$ is separable.
Since the trace is basis independent, the entanglement content will also be basis independent. This obviously also includes the case where the entanglement is maximal.
As an aside, there are a lot more entanglement measures you can define, each with there own properties. E.g., you can also look into entanglement entropy if you're curious.
Let $\mathcal{H}$ with dimension $\dim \mathcal{H} = 2^n$ where $n \in \mathbb{N}$ be a Hilbert space describing your system of interest. This could be a Hilbert space for a system of $n$ qubits, for instance. There are two choices to be made.
Entanglement is emphatically dependent on the choice of tensor product factorization. For instance, if we choose the tensor product factorization $$\mathcal{H} \cong \mathbb{C}^{2^n}$$ entanglement cannot even be defined between subsytems because there are no subsystems! But, if we just move into another tensor product factorization $$\mathcal{H} \cong \mathbb{C}^2 \otimes \mathbb{C}^2 \otimes ... \otimes \mathbb{C}^2$$ there are now subsystems and entanglement can exist between them.
Entanglement entropy is not dependent on the choice of basis for the tensor factors. This follows, as others remark, from the fact that matrix trace is cyclic and so invariant under conjugating the argument by unitary matrices $$\text{Tr}(U M U^\dagger) = \text{Tr}(MU^\dagger U) = \text{Tr}(M).$$ This is actually a fundamental property of any entanglement entropy measure, so anything that you call entanglement entropy must be independent of choice of individual tensor factor basis. This is obvious after understanding that entanglement entropy is a quantity defined between subsystems, so it should only depend on the choice of labelling subsystems. And, the mathematical structure corresponding to labelling subsystems is precisely a choice of tensor product factorization.
Hence, the answer to your question is yes. This follows immediately from the more general fact that entanglement entropy is independent of the choice of individual tensor factor basis you choose to compute it in.
[1] In actuality, some tensor product factorizations are equivalent to other tensor product factorizations in the sense that the entanglement entropy in one tensor product factorization is the same as in another, equivalent tensor product factorization. Then, the correct object to consider is a tensor product structure, which is an equivalence class of tensor product factorizations. The precise definition of a tensor product structure is given on page 6 of the paper "Locality from the spectrum".
