Why does current density have a direction and not current?
According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$.
The current is defined according to a surface $A_t$ (and is a local quantity: the position of the surface). Since the surface may be tilted (not perpendicular to $\vec{n_d}$), then in general, we should write $$I=n\;q \; \vec{A_t}.\vec{v_d}=n\;q\;A_t \; \vec{n_{A_t}}.\vec{v_d}=n\;q\;A_t \; cos(\theta).v_d$$ where $\theta$ is the angle between $\vec{A_t}$ and $\vec{v_d}$ (If the charge $q$ is negative, $\vec{I}$ and $\vec{n_d}$ would have opposite orientations, which fits with the usual convention of the electric current). Of course, if we have considered a surface $A$ which is perpendicular to the drift, the current would be the same but we would write $$I=n\;q \; \vec{A}.\vec{v_d}=n\;q\;A \; v_d$$
That is, let's talk about the densities. We have $$\mathrm{d}I=n\;q\;v_d\;\mathrm{d}A=n\;q\;v_d\;cos(\theta)\;\mathrm{d}A_t$$ The scalar current density is given by $$j=\frac{\mathrm{d}I}{\mathrm{d}A}=n\;q\;v_d=\frac{1}{cos(\theta)}\frac{\mathrm{d}I}{\mathrm{d}A_t}$$ Yet, you see that you have to be careful about the differential area that you put in the denominator. The current could then be calculated as $$I=\int_A j\;\mathrm{d}A=\int_{A_t} j\;cos(\theta)\;\mathrm{d}A_t$$ Here also, we see a possible source of confusion. This can be fixed if we define a vector current density as $$\vec{j}=j\;\vec{n_d}=n\;q\;v_d\;\vec{n_d}$$ The direction of $\vec{j}$ is resolved too: it's that of the local drift in the considered material position. That direction may be different than that of the average drift of the whole current $\vec{I}$. That is, the expression of the current can be written simply as $$I=\int_A \vec{j}\;\mathrm{d}\vec{A}=\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}$$ Or, you could use your own convention and write $$\vec{I}=\bigg(\int_A \vec{j}\;\mathrm{d}\vec{A}\bigg)\;\vec{n_d}=\bigg(\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}\bigg)\;\vec{n_d}$$
As you already mentioned $$ I = \int \vec{J} \cdot d\vec{A} $$ current is the charge-flow through a given surface. So If one talks about currents, the surface (and therefore its normal direction) is to be understood beforehand.
You could assign a direction like this: $$ \frac{\vec{I}}{I} = \frac{\int d\vec{A}}{A} $$ which is the mean normal direction of the surface but this makes no sense for curved surfaces (e.g $\vec{I} = 0$ for a ball, regardless of the magnitude $I$).
This is basically just rephrasing some of the existing answers, but:
Current is a scalar $I$ with units of [J/s]. It is defined as $I = dQ / dt$.
Not quite - $dQ/dt$ isn't precisely defined. What is $Q$ the charge of? Current is actually defined to be the charge per time passing through some surface $S$. In terms of current density it can be expressed as $I := \iint_S {\bf J} \cdot d{\bf S}$. The fact that it is only defined with respect to some surface $S$ means that it is an inherently global quantity, unlike current density ${\bf J}$, which can be unambiguously defined at a single point. (Current is also in general not a very physical quantity, unlike current density, because in principle $S$ can be any crazy Gaussian surface. It's exactly analagous to the difference between electric field vs. electric flux.) As others have mentioned, if the surface $S$ is curved then your proposal to integrate with respect to $dS$ instead of $d{\bf S}$ doesn't work.
The problem is that we almost always consider the current going through thin wires, in which case none of these subtleties come up. For the mathematical intuition, it's better to think of current going through a bulk conductor (possibly including eddy currents, etc.) instead.