This is nicely understood by duality (in the sense of dualizable objects in monoidal categories) in the monoidal category KK. The relevant results are collected and referenced on the nLab at Poincaré duality algebra:
For $X$ a closed manifold equipped with a twist $\chi$ (the class of a circle 2-bundle, realized as a $U(1)$bundle gerbe etc., let $C_\chi(X)$ be the corresponding twisted groupoid convolution C*-algebra, well defined up to Morita equivalence and hence KK-equivalence. This is the $C^\ast$-algebra such that
$$
KK_\bullet(\mathbb{C}, C_\chi(X)) \simeq K^{\bullet+\chi}(X)
$$
is the $\chi$-twisted K-theory of $X$.
Then here is the statement: The dual object of $C_\chi(X)$ in $(KK, \otimes)$ is
$$
\left(
C_\chi\left(X\right)
\right)^\vee
\simeq
C_{-\chi - W_3(T X)}(X)
\,,
$$
where $W_3(T Q)$ is the third integral Stiefel-Whitney class of the tangent bundle of $X$.
This formula implies everything in this business.
For instance it implies that $C_\chi(X)$ is self-dual, and hence (since duality in KK is duality between K-homology and K-cohomology) that $X$ exhibits K-Poincaré duality, precisely if $\chi = 0$ and $W_3(T X) = 0$. The latter is precisely the condition of spin^c structure.
But you get much more by playing with the formula, and this is I think what you are after. For instance given a map of closed manifolds $i \colon Q \to X$ with $X$ carrying a twist $\chi$ and $Q$ carrying the twist $i^\ast \chi$, we get the corresponding pullback map of twisted convolution algebras
$$
i^\ast \colon C_\chi(X) \to C_{i^\ast \chi}(Q)
\,.
$$
Now since both objects have dualty, we can form the corresponding dual morphism by the general formula in monoidal categories:
$$
i_! := (i^\ast)^\vee : (C_{i^\ast \chi}(Q))^\vee \to (C_\chi(X))^\vee
\,,
$$
hence by the above
$$
i_! : C_{-i^\ast \chi - W_3(T Q)}(Q) \to C_{-\chi-W_3(T X)}(X)
\,.
$$
By just relabelling the original twist for transparency as $\chi \mapsto - \chi - W_3(T X)$ this is equivalently a morphism
$$
i_! : C_{i^\ast \chi + W_3(N_i Q)}(Q) \to C_{\chi}(X)
\,.
$$
(Here $N_i Q$ denotes the normal bundle of $Q$ in $X$ via $i$.)
By postcomposition in KK and using the general relation that $KK(\mathbb{C}, A) \simeq K(A)$ this yields a homomorphism
$$
i_! : K^{\bullet + i^\ast \chi + W_3(N_i Q)} \to K^{\bullet + \chi}(X)
$$
from the $(i^\ast \chi + W_3(N_i Q))$-twisted K-theory of $Q$ to the $\chi$-twisted K-thory of $X$.
This is the general twisted Umkehr map in twisted K-theory. Notice again that it is simply the dual morphism of $i^\ast$ in $(KK,\otimes)$.
If now the normal bundle has $Spin^c$-structure then that term drops out, and so on. But in general it is like this.
If here we think of $X$ as the target spacetime for a type II superstring, of $\chi$ as the instanton sector of the B-field, and of $Q \hookrightarrow X$ as the worldvolume of a D-brane, then the $(i^\ast \chi + W_3(N_i Q))$-twisted K-cocycles on $Q$ are the (underlying instanton sectors of) the Chan-Paton gauge fields subject to the
Freed-Witten-Kapustin anomaly cancellation condition and their image under the above twisted Umkehr map is the D_brane charge of the D-brane with that Chan-Paton bundle.
All this just means really: the dual map of $i^\ast$ picks up twists as above. See at Poincaré duality algebra for citations. All these results go through also for $G$-equivariant twisted K-theory.
If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $\mathbb{R}^n$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $k+1 \ge 3$ ends.
Freudenthal and Hopf proved that a finitely generated group has $0$, $1$, $2$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption.
I suspect (but am really not sure) that it may be possible to extend the argument to closed $n$-manifolds with universal cover only homotopy equivalent to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$. Perhaps it can be shown that the universal cover must have $k+1$ ends by thinking of the separation properties of representatives of $H_{n-1}(\bigvee_i S^{n-1})$.
Best Answer
No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.
The reason is simply that there are finitely presented groups which have no nontrivial finite quotient. One example is Higman's group $H$, see https://en.wikipedia.org/wiki/Higman_group.
The key features of $H$ are:
The proof of 1,2 can be found in Tao's blog https://terrytao.wordpress.com/2008/10/06/finite-subsets-of-groups-with-no-finite-models/, and the proof of 3,4 in ''The topology of discrete groups'' by Baumslag, Dyer, Heller).
Now pick an element $1 \neq x \in H$, which induces an injective homomorphism $\mathbb{Z} \to H$ and form the amalgamated product $G=H \ast_{\mathbb{Z}} H$. The group $G$ is infinite and has no nontrivial homomorphism to a finite group $F$, since any homomorphism $G \to F$ must vanish on the two copies of $H$.
The pushout $BH \cup_{S^1} BH$ is aspherical by Whiteheads asphericity theorem and hence a model for $BG$. I have designed things so that $H_2(BG) \cong \mathbb{Z}$ and all other homology groups are trivial. In particular, $G$ is perfect, and the Quillen plus construction $BG^+$ must be homotopy equivalent to $S^2$, so that there is a homology equivalence $f:BG \to S^2$. Now let $V \to S^2$ be the nontrivial oriented vector bundle of rank $d$, which has $w_2 (V) \neq 0$. It follows that the vector bundle $f^\ast V \to BG$ is not spin. $BG$ has no nontrivial cover, and $BG$ is aspherical, so the pullback of $f^\ast V$ to the universal cover is trivial.
Now there exists, when $d \geq 4$, a closed $d$-manifold $M$ with a $2$-connected map $\ell: M \to BG$ and a bundle isomorphism $TM\oplus \mathbb{R}\cong \ell^\ast f^\ast V \oplus \mathbb{R}$. This is achieved by surgery below the middle dimension. In particular, $\pi_1 (M)\cong G$. Hence $\pi_1(M)$ has no nontrivial finite index normal subgroup, and therefore no nontrivial finite index subgroup at all. It follows that $M$ does not have a nontrivial finite cover.
By construction, $w_2 (TM) \neq 0$, but the universal cover of $M$ is stably parallelizable.