One example arises as the total space of a family of Godeaux surfaces over $\mathbb{P}^1_k$ with sufficiently positive discriminant.

Let $k$ be $\mathbb{C}$ (because that is fewer keystrokes in LaTeX).  Let $\mathbb{P}^1_k$ be $\text{Proj}\ k[R,S]$.  Let $\mathbb{P}^3_k$ be $\text{Proj}\ k[T_0,T_1,T_2,T_3]$.  Let $e$ be a nonnegative integer.  Let $b_0,b_1,b_2,b_3 \in k[R,S]_e$ be a $4$-tuple of homogeneous polynomials of degree $e$ on $\mathbb{P}^1_k$.

<B>Hypothesis 1.</B> The integer $e$ is $\geq 3$.  Every zero in $\mathbb{P}^1_k$ of each polynomial $b_i$ is a simple zero, and there is no simultaneous zero of two or more of the polynomials $b_i,b_j$.

<B>Notation 2.</B> Denote by $f$ be the section of $\text{pr}_{\mathbb{P}^1}^*\mathcal{O}(e)\otimes \text{pr}_{\mathbb{P}^3}^*\mathcal{O}(5)$, $$f = b_0 T_0^5 + b_1 T_1^5 + b_2 T_2^5 + b_3 T_3^5.$$  Denote by $Y$ be the zero scheme of $f$ as a hypersurface in $\mathbb{P}^1_k\times_{\text{Spec}\ k} \mathbb{P}^3_k$.  For each $i=0,1,2,3$, denote by $p_i$ the $k$-point of $\mathbb{P}^3_k$ where $T_j$ vanishes for every $j\neq i$.  Also denote by $Y_i\subset Y$ the product $\text{Zero}(b_i)\times\{p_i\}$.  Finally, denote the restrictions of $\text{pr}_{\mathbb{P}^1}$ and $\text{pr}_{\mathbb{P}^3}$ to $Y$ as follows, $$\pi:Y\to \mathbb{P}^1_k, \ \ \rho:Y\to \mathbb{P}^3_k.$$

<B>Lemma 3.</B>  The hypersurface $Y$ is smooth.  The singular locus of the projection $\pi$ equals $Y_0\cup Y_1\cup Y_2 \cup Y_3$.  The projection $\rho$ is smooth over $p_i$ for every $i=0,1,2,3$.  The dualizing sheaf of $Y$ equals the restriction of $\text{pr}_{\mathbb{P}^1}^*\mathcal{O}(e-2)\otimes \text{pr}_{\mathbb{P}^3}^*\mathcal{O}(1)$. 

<B>Proof.</B>  By the Jacobian criterion, the projection $\pi$ is smooth away from $\cup_i Y_i$.  The fiber of $\rho$ over $p_i$ is the zero scheme of $b_i$, and this is reduced (hence smooth) by hypothesis.  Thus, the projection morphism $\rho$ is smooth at every point of $\cup_i Y_i$.  For every point of $Y$, either $\pi$ or $\rho$ is smooth, and hence $Y$ is everywhere smooth.  The remaining computations are straightforward. <B>QED</B>

Let $\mu_5$ denote the group of $5^{\text{th}}$ roots of unity in $\mathbb{G}_{m,k}$.  Let $\mu_5$ act on $\mathbb{P}^1_k\times_{\text{Spec}\ k}\mathbb{P}^3_k$ as follows, $$\zeta \bullet([S_0,S_1],[T_0,T_1,T_2,T_3]) = ([S_0,S_1],[T_0,\zeta T_1, \zeta^2 T_2, \zeta^3 T_3]).$$  The polynomial $f$ is invariant for this action. Thus, there is an induced action on $Y$.

<B>Notation 4.</B> Denote the quotient of this $\mu_5$-action on $Y$ by $\nu:Y\to X'$.

<B>Lemma 5.</B> The action of $\mu_5$ on $Y\setminus \cup_i Y_i$ is free.  For every point of $\cup_i Y_i$, the <I>age</I> (in the sense of Reid -- Shepherd-Barron -- Tai) is $> 1$.

<B>Proof.</B>  The fixed points of the action on $\mathbb{P}^3_k$ are the points $p_i$.  Since the projection $\rho$ is equivariant, every fixed point in $Y$ maps to $p_i$ for some $i$.  Thus, the fixed locus is contained in $\cup_i Y_i$.

Since the projection $\rho$ is &eacute;tale at every point of $\cup_i Y_i$, the age of that fixed point in $Y$ equals the age of the image fixed point, say $p_i$, in $\mathbb{P}^3_k$.  By the choice of action of $\mu_5$ on $\mathbb{P}^3_k$, every eigenvalue that occurs in $T_{\mathbb{P}^3,p_i}$ has multiplicity $1$.  Moreover, since the fixed points are isolated, none of these eigenvalues equals $1$.  Thus, for each choice of primitive generator $\zeta$ of $\mu_5$, the eigenvalues $\zeta^a,\zeta^b,\zeta_c$ that occur have $0<a<b<c <5$, up to rearrangement.  Thus, the sum $a+b+c$ is at least $1+2+3=6$.  Therefore the age $\alpha(\mathbb{P}^3,p_i) = (a+b+c)/5$ is strictly larger than $1$. <B>QED</B> 

<B>Proposition 6.</B> The $k$-scheme $X'$ is projective, normal, $\mathbb{Q}$-Gorenstein, has ample $\mathbb{Q}$-canonical divisor class, and has only terminal finite quotient singularities.  Finally, $X'$ is simply connected.  Thus, every projective desingularization of $X'$ is a simply connected, projective $3$-fold of general type.

<B>Proof.</B>  The only nontrivial claim is that $X'$ has only terminal singularities.  This follows by the Reid--Shepherd-Barron--Tai criterion and the computation of the age above. 
Since $Y$ is simply connected, also $X'$ is simply connected.
<B>QED</B>

Since the morphism $\pi$ is $\mu_5$-invariant, there is a unique $k$-morphism, $$\pi':X'\to \mathbb{P}^1_k,$$ such that $\pi'\circ \nu$ equals $\pi$.  By Lemma 3, the singular locus of $\pi'$ equals $X'_0\cup X'_1\cup X'_2 \cup X'_3$, for $X'_i:= \nu(Y_i)$.  Thus, the smooth locus of $\pi$ equals the open complement $U$ of $\cup_i X'_i$ in $X'$.  On $U$, the transitivity exact sequence of $\pi$ gives a short exact sequence, $$0 \to \pi^*\Omega_{\mathbb{P}^1/k}^1 \to \Omega_{U/k} \to \Omega_{\pi} \to 0.$$  This induces a short exact sequence, $$0 \to \pi^*\Omega_{\mathbb{P}^1/k}\otimes\Omega_{\pi} \to \Omega^2_{U/k} \to \Omega^2_{\pi} \to 0.$$  This also induces an isomorphism, $$\Omega^3_{U/k} \cong \pi^*\Omega_{\mathbb{P}^1/k}\otimes \Omega^2_{\pi}.$$

<B>Proposition 7.</B>  For $r=1,2$, the only global section of $\Omega^r_\pi$ on $U$ is the zero section.  For $r=1,2,3$, the only global section of $\Omega^r_{U/k}$ on $U$ is the zero section.  Thus, for every projective desingularization $X$ of $X'$, the only global section of $\Omega^r_{X/k}$ on $X$ is the zero section.

<B>Proof.</B>  These coherent sheaves are locally free sheaves on a smooth $k$-scheme.  Every nonzero section on $U$ restricts to a nonzero section of the restriction on the generic fiber $F$ of $\pi$.  The restriction of $\Omega^r_\pi$ to the generic fiber $F$ equals $\Omega^r_{F/k(\mathbb{P}^1)}$. The generic fiber $F$ of $\pi$ is a Godeaux surface over $k(\mathbb{P}^1)$.  Thus, $\Omega^r_{F/k(\mathbb{P}^1)}$ has only the zero global section for $r=1$ and $r=2$.  

Since the global sections of $\Omega^r_\pi$ are zero for $r>0$, the only possibility for a nonzero global section of $\Omega^r_{U/k}$ is a global section of $\pi^*\Omega_{\mathbb{P}^1_k/k}$ when $r=1$.  Of course this equals the restriction to $U$ of $\pi^*\mathcal{O}(-2)$.  Since $X'$ is normal and since $\cup_i X'_i$ consists of finitely many points, every global section of $\pi^*\mathcal{O}(-2)$ on $U$ is the restriction of a unique global section on all of $X'$ of $\pi^*\mathcal{O}(-2)$.  By the projection formula, every such global section is the zero global section. <B>QED</B>