$X = \Bbb{CP}^4 \#\Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.
Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $\Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;\Bbb Z)$ as $x_1, x_2$.
1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.
2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.
3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $\Bbb{CP}^2 \vee \Bbb{CP}^2$. Because $X_4 \hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X \to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f \vee f$, where $f$ is the restriction of the classifying map of $\Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(\Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.
Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2\ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).
Then because $\chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+\ell n) + 16(m+n) - (2k+1)^4 - (2\ell+1)^4 + 10(2k+1)^2 + 10(2\ell+1)^2 - 50.$$
Simplifying we get $$80 = 32(km + \ell n - k^3 - \ell^3 + k^2 + \ell^2) + 16(k+\ell - k^4 - \ell^4 +m +n)$$
Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 \equiv 16(k+\ell - k^4- \ell^4) \mod 32$$ But $k+\ell$ is odd iff $k^4 + \ell^4$ is odd, so the right side is $0 \mod 32$. This is a contradiction, as desired.
That this was so much work suggests that this is, uh, the wrong approach to prove that $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.
I am a little late to the game here, but I wanted to share some pictures, which might help intuition. Paul Frost answers your question, but I thought of this a different way. Take the standard $x,y$-plane in 3-space. This is an unbounded surface. If I were to add a handle to it, it is still unbounded. The only question is whether this handle makes the surface non-orientable or not. In the image below, I have both ways I can attach the handle. These surfaces are topologically a punctured torus and a punctured Klein bottle, but drawn this way, I think it sheds light on what you were thinking about. The orange circle in the bottom is the circle of immersion necessary to put the Klein in 3-space. And while there is obviously nontrivial loops, these are unbounded. Last is an example of a non-orientable surface with boundary, but it is embedded.
Best Answer
Every complex manifold is orientable, as every complex vector space as a canonical orientation as a real space. Namely if $V$, is a complex vector space, and $B= (u_1,...,u_n)$ is a base (over C), then $B^*=(u_1,...,u_n, iu_1,...iu_n)$ is a base over $\bf R$. Note that if $B'$ is another base over $C$ and $l$ the unique linear map $C$ map such that $lB=B'$, $lB^*=B'^*$, and the determinant of $l$, viwed as a $R$ linear map is the squre of the modulus of the determinant of $l$, hence positive. Thus the orientation given by $B$ is the same than that given by $B'$, and complex linear maps preserves this orientation.